Division of polynomials

Class 11 Mathematics, unit 5, book pages 150–161

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A polynomial can be divided by another polynomial just as one whole number is divided by another, leaving a quotient and a remainder. This unit sorts algebraic expressions into polynomial, rational and irrational kinds, shows long division of polynomials, and then proves two shortcuts: the Remainder Theorem, which finds the remainder without dividing, and the Factor Theorem, which tells when is a factor. These are used to find zeros, to factorise cubic polynomials and to solve problems about volumes and areas.

Learning outcomes (5)
  • Divide a polynomial of degree up to 4 by a linear and quadratic polynomial to identify quotient and remainder.
  • Recognize remainder theorem and factor theorem.
  • Demonstrate and apply remainder theorem.
  • Analyze and apply factor theorem to factorize a cubic polynomial.
  • Apply concepts of remainder and factor theorem to real world problems.

5.1Algebraic expressions and polynomials

An algebraic expression is a statement in which variables or constants or both are connected by the arithmetic operations . Examples from the book:

5.1.1 Kinds of algebraic expressions

Algebraic expressions are of three kinds:

  1. Polynomial expressions
  2. Rational expressions
  3. Irrational expressions
1. Polynomial expressions (polynomials)

Polynomials are algebraic expressions of one or more terms in which the exponents of the variables are whole numbers. The general form of a polynomial in is

Examples: .

Not every expression is a polynomial. The following are not polynomials because an exponent is not a whole number (a positive integer or zero):

ExpressionWhy it is not a polynomial
exponent is negative
has a negative exponent
, exponent is a fraction
exponent is a fraction
Types of polynomials with respect to degree
TypeMeaningExamples
Zero polynomial (no degree)the polynomial ; all its coefficients are zero, so no degree is assigned, and
Constant polynomialdegree zero
Linear polynomialdegree one
Quadratic polynomialdegree two
Cubic polynomialdegree three
Bi-quadratic polynomialdegree four

Polynomials of higher degree are called degree five and so on.

2. Rational expressions

An algebraic expression of the form , where and are polynomials and (that is, is not the zero polynomial), is called a rational expression.

Examples: .

3. Irrational expressions

An algebraic expression which cannot be expressed in the form with and polynomials ( not the zero polynomial) is called an irrational expression.

Examples: .

5.2Division of a polynomialOften examined

To divide one polynomial by another we use long division, exactly as in arithmetic. The book explains it with an example: divide by .

Long division by a quadratic
Divide by .
  1. The dividend has no term and no term, so insert them with coefficient .
  2. Divide the leading terms: . This is the first term of the quotient. Multiply the divisor by and subtract.
  3. Divide the new leading terms: . Multiply the divisor by and subtract.
  4. Divide again: . Multiply the divisor by and subtract.
  5. The remainder has degree 1, which is less than the degree 2 of the divisor, so the division stops.
  6. Check with dividend quotient divisor remainder.

5.3Remainder theoremOften examined

To see the idea, divide a polynomial by and compare the remainder with the value of the polynomial at .

Example 1: dividing and comparing
If and , divide by and find the quotient and remainder.
  1. Divide the leading terms: . Multiply by and subtract.
  2. Divide: . Multiply by and subtract.
  3. Divide: . Multiply by and subtract.
  4. A constant has lower degree than , so we stop.
  5. Check: quotient divisor remainder gives back the dividend.
  6. Now put (the value that makes the divisor zero) in the dividend.

So is the same number as the remainder obtained by long division. We deduce: when is divided by the remainder is , and the value of the dividend at also equals .

Derivation Proof of the Remainder Theorem

Aim. Show that when is divided by the remainder is .

Assumptions.
  • is a constant
  • is the quotient and the remainder when is divided by
  1. Use the division rule: dividend quotient divisor remainder. The divisor is linear, so the remainder has degree less than 1, that is, it is a constant .
  2. Equation (i) is true for every , so substitute in it.
  3. The first term vanishes because .
Result

The remainder on dividing by is just the value of the polynomial at , so it can be found by substitution, without long division.

Example 2
Find the remainder if is divided by .
  1. Name the polynomial and the divisor.
  2. By the Remainder Theorem the remainder is , since gives .
  3. Substitute .
Example 3
Find the value of when is divided by and the remainder is .
  1. Let and . The divisor is zero at , so the remainder is .
  2. Substitute .
  3. The remainder is given as 4, so put and simplify.
  4. Solve for .
Worked example: remainder without dividing
Find the remainder when is divided by , without dividing. Then check by long division.
  1. Write the divisor in the form . Since , we have , and by the Remainder Theorem the remainder is .
  2. Substitute in . Take care with the signs: .
  3. Check. Long division (inserting ) gives quotient . Multiply it by the divisor and add the remainder; we must get back .

5.3.1Zeros of a polynomialOften examined

Consider the equation 2x+5=9. Subtracting 5 from both sides gives 2x=4, so x=2. The number 2 is called a root of the equation because it satisfies it. In the same way, the roots of a polynomial p(x) are the values of x that satisfy p(x)=0. These roots are called the zeros of the polynomial.

Zero of a polynomial
Let P(x) be any polynomial. A number a is called a zero of P(x) if and only if P(a)=0.

For example, 5 and -5 are the zeros of p(x)=x^{2}-25, because

p(5)=(5)^{2}-25=25-25=0,\qquad p(-5)=(-5)^{2}-25=25-25=0
The graph of p(x)=x^{2}-25 crosses the x-axis exactly at its zeros x=5 and x=-5.

So finding the zeros of a polynomial means finding those values of x which make the value of the polynomial equal to zero.

Example 4
Is -3 a zero of the polynomial p(x)=2x^{4}+7x^{3}-4x^{2}-27x-18?
  1. -3 is a zero of p(x) if and only if p(-3)=0, so evaluate p(-3).
    p(-3)=2(-3)^{4}+7(-3)^{3}-4(-3)^{2}-27(-3)-18
  2. Work out each power.
    =2(81)+7(-27)-4(9)-27(-3)-18
  3. Multiply.
    =162-189-36+81-18
  4. Add the positive and the negative terms separately.
    =(162+81)-(189+36+18)=243-243=0
  5. Since p(-3)=0, the number -3 is a zero of p(x).
p(-3)=0,\ \text{so }-3\text{ is a zero}
Example 5
If the zeros of a polynomial are 0,\ 6,\ -1, find the polynomial.
  1. If a is a zero then x-a is a factor. Let the required polynomial be g(x) and use the zeros x=0,6,-1.
    g(x)=(x-0)(x-6)(x+1)
  2. Multiply the last two brackets first.
    (x-6)(x+1)=x^{2}+x-6x-6=x^{2}-5x-6
  3. Multiply by x.
    g(x)=x(x^{2}-5x-6)=x^{3}-5x^{2}-6x
g(x)=x^{3}-5x^{2}-6x
Example 6
If one zero of g(x)=2x^{3}+x^{2}-2x-1 is -\dfrac12, find its other zeros.
  1. If -\tfrac12 is a zero, then x=-\tfrac12 gives 2x+1=0, so 2x+1 is a factor of g(x). The other zeros come from factorising the rest.
    x=-\dfrac12\;\Longrightarrow\;2x+1\text{ is a factor}
  2. Divide g(x) by 2x+1. First term: 2x^{3}\div2x=x^{2}\; subtract x^{2}(2x+1)=2x^{3}+x^{2}.
    (2x^{3}+x^{2}-2x-1)-(2x^{3}+x^{2})=-2x-1
  3. Next term: -2x\div2x=-1\; subtract -1(2x+1)=-2x-1.
    (-2x-1)-(-2x-1)=0
  4. The remainder is 0 and the quotient is x^{2}-1.
    2x^{3}+x^{2}-2x-1=(2x+1)(x^{2}-1)
  5. Factorise x^{2}-1 as a difference of squares.
    =(2x+1)(x-1)(x+1)
  6. Set each new factor equal to zero.
    x-1=0\Rightarrow x=1\;\qquad x+1=0\Rightarrow x=-1
\text{Other zeros: }1\text{ and }-1
Worked example: finding every zero of a cubic and seeing them on its graph
Find all the zeros of p(x)=x^{3}-3x^{2}-4x+12.
  1. A whole-number zero must divide the last term 12. Try small values with the Factor Theorem.
    p(1)=1-3-4+12=6\neq0
  2. Try x=2.
    p(2)=8-12-8+12=0
  3. So 2 is a zero and x-2 is a factor. Divide to get the quadratic factor: x^{3}\div x=x^{2}, then -x^{2}\div x=-x, then -6x\div x=-6, with remainder 0.
    x^{3}-3x^{2}-4x+12=(x-2)(x^{2}-x-6)
  4. Factorise the quadratic: two numbers with product -6 and sum -1 are -3 and 2.
    x^{2}-x-6=(x-3)(x+2)
  5. Each linear factor gives one zero.
    p(x)=(x-2)(x-3)(x+2)=0\ \Rightarrow\ x=2,\ 3,\ -2
  6. Check one of the new zeros: p(3)=27-27-12+12=0. The graph below cuts the x-axis at exactly these three points.
The graph of p(x)=x^{3}-3x^{2}-4x+12 meets the x-axis at -2, 2 and 3, the three zeros found by trial and division.
\text{zeros: }-2,\ 2,\ 3

5.4Factor theoremOften examined

The Factor Theorem is a result of the Remainder Theorem and is based on the same reasoning.

Derivation Proof of the Factor Theorem

Aim. Show that x-c is a factor of p(x) if and only if p(c)=0. 'If and only if' means we must prove both directions.

Assumptions.
  • q(x) is the quotient and r the remainder when p(x) is divided by x-c
  1. Write the division rule: dividend = quotient \times divisor + remainder.
    p(x)=(x-c)\,q(x)+r\qquad(\text{i})
  2. First direction. Suppose x-c is a factor of p(x). A factor divides completely, so the remainder is 0.
    r=0
  3. By the Remainder Theorem the remainder is p(c), so p(c)=r=0.
    r=p(c)\;\Longrightarrow\;p(c)=0
  4. Converse. Suppose p(c)=0. By the Remainder Theorem the remainder is r=p(c)=0.
    r=p(c)=0
  5. Then result (i) reduces to a product with no remainder.
    p(x)=(x-c)\,q(x)
  6. So x-c divides p(x) exactly, that is, x-c is a factor of p(x).
Result(x-c)\text{ is a factor of }p(x)\iff p(c)=0

To test whether x-c is a factor, substitute x=c: a zero value means it is a factor, a non-zero value means it is not (and that value is the remainder).

Example 7
Show that y-1 is a factor of y^{4}-24y^{2}-13y+36.
  1. Let f(y)=y^{4}-24y^{2}-13y+36. By the Factor Theorem y-1 is a factor of f(y) if f(1)=0, so find f(1).
    f(y)=y^{4}-24y^{2}-13y+36
  2. Substitute y=1.
    f(1)=(1)^{4}-24(1)^{2}-13(1)+36=1-24-13+36
  3. Add the terms.
    =37-37=0
  4. The remainder is 0, so y-1 is a factor of y^{4}-24y^{2}-13y+36.
f(1)=0,\ \text{so }y-1\text{ is a factor}
Example 8
Determine the value of k for which x+3 is a factor of (x+2)^{5}+(3x+k).
  1. Let f(x)=(x+2)^{5}+(3x+k) and d(x)=x+3. As x+3 is a factor, the Factor Theorem gives f(-3)=0.
    f(-3)=0
  2. Substitute x=-3.
    (-3+2)^{5}+[3(-3)+k]=0
  3. Simplify: (-1)^{5}=-1.
    (-1)^{5}+(-9+k)=0\;\Longrightarrow\;-1-9+k=0
  4. Solve for k.
    -10+k=0\;\Longrightarrow\;k=10
  5. For k=10, x+3 is a factor of (x+2)^{5}+(3x+k).
k=10
Worked example: two given factors fix two unknown coefficients
If x-1 and x+2 are factors of p(x)=x^{3}+ax^{2}+bx-2, find a and b, and the third factor.
  1. A factor x-c means p(c)=0 (Factor Theorem). Use x-1, so c=1.
    p(1)=1+a+b-2=0\ \Rightarrow\ a+b=1
  2. Use x+2=x-(-2), so c=-2.
    p(-2)=-8+4a-2b-2=0\ \Rightarrow\ 2a-b=5
  3. Add the two equations to remove b.
    3a=6\ \Rightarrow\ a=2,\quad b=1-2=-1
  4. So the polynomial is known. The product of the two given factors is (x-1)(x+2)=x^{2}+x-2. Divide p(x) by it: x^{3}\div x^{2}=x, subtract x(x^{2}+x-2), leaving x^{2}+x-2; then divide by x^{2} to get +1, remainder 0.
    x^{3}+2x^{2}-x-2=(x^{2}+x-2)(x+1)
  5. The third factor is x+1, so p(x)=(x-1)(x+1)(x+2).
The graph of p(x)=x^{3}+2x^{2}-x-2 crosses the x-axis at -2, -1 and 1, one point for each factor x+2, x+1, x-1.
a=2,\ b=-1;\quad \text{third factor }x+1

5.5Factorization of a cubic polynomialOften examined

With the help of the Factor Theorem together with some intelligent guessing, we can factorise polynomials of higher degree. The book explains the process on the cubic polynomial

f(x)=x^{3}-2x^{2}-5x+6
Step I: find one factor by trial

Substitute \pm1,\pm2,\pm3 and \pm4 to trace a root, using the Factor Theorem. First try x-1: it is a factor if f(1)=0.

f(1)=(1)^{3}-2(1)^{2}-5(1)+6=1-2-5+6=7-7=0

Hence x-1 is a factor of f(x).

Step II: divide to get the other factor

Divide x^{3}-2x^{2}-5x+6 by x-1 to find the other factor q(x).

  1. x^{3}\div x=x^{2}. Subtract x^{2}(x-1)=x^{3}-x^{2}.
    (x^{3}-2x^{2}-5x+6)-(x^{3}-x^{2})=-x^{2}-5x+6
  2. -x^{2}\div x=-x. Subtract -x(x-1)=-x^{2}+x.
    (-x^{2}-5x+6)-(-x^{2}+x)=-6x+6
  3. -6x\div x=-6. Subtract -6(x-1)=-6x+6.
    (-6x+6)-(-6x+6)=0
  4. The remainder is 0, as the Factor Theorem promised.
    x^{3}-2x^{2}-5x+6=(x-1)(x^{2}-x-6)
Step III: factorise the quadratic factor

Factorise the quadratic factor x^{2}-x-6 for the other linear factors. Split the middle term -x into -3x+2x (numbers with product -6 and sum -1).

  1. Replace -x by -3x+2x.
    x^{3}-2x^{2}-5x+6=(x-1)(x^{2}-x-6)=(x-1)(x^{2}-3x+2x-6)
  2. Group in pairs and take out the common factor of each pair.
    =(x-1)\big[x(x-3)+2(x-3)\big]
  3. Take out the common bracket (x-3).
    =(x-1)(x-3)(x+2)
x^{3}-2x^{2}-5x+6=(x-1)(x-3)(x+2)
Example 9
If two linear factors of the polynomial 2y^{3}+y^{2}-8y-4 are (2y+1) and (y-2), find its third factor.

The book first shows three wrong attempts at guessing the third factor, by writing 2y^{3}+y^{2}-8y-4=(2y+1)(y-2)\,(\ ?\ ) and expanding. Guessing is unreliable, so find the third factor by division instead.

  1. The product of the two known factors is a quadratic factor of the cubic.
    (2y+1)(y-2)=2y^{2}-4y+y-2=2y^{2}-3y-2
  2. Divide 2y^{3}+y^{2}-8y-4 by 2y^{2}-3y-2. First term: 2y^{3}\div2y^{2}=y. Subtract y(2y^{2}-3y-2)=2y^{3}-3y^{2}-2y.
    (2y^{3}+y^{2}-8y-4)-(2y^{3}-3y^{2}-2y)=4y^{2}-6y-4
  3. Next term: 4y^{2}\div2y^{2}=2. Subtract 2(2y^{2}-3y-2)=4y^{2}-6y-4.
    (4y^{2}-6y-4)-(4y^{2}-6y-4)=0
  4. The remainder is 0, so the quotient y+2 is the missing factor.
    2y^{3}+y^{2}-8y-4=(2y+1)(y-2)(y+2)
  5. Check with the last-term rule: the constants of the three factors multiply to the last term.
    (+1)(-2)(+2)=-4
\text{Third factor }(y+2)
Example 10
Factorise x^{3}-5x-2 by the Factor Theorem.
  1. Let f(x)=x^{3}-5x-2. The last term is -2, so try numbers that divide 2, for example x=-2.
    f(x)=x^{3}-5x-2
  2. Evaluate f(-2).
    f(-2)=(-2)^{3}-5(-2)-2=-8+10-2=0
  3. Hence x+2 is one of the factors of x^{3}-5x-2.
  4. Divide x^{3}-5x-2 by x+2 (insert 0x^{2}). x^{3}\div x=x^{2}; subtract x^{2}(x+2)=x^{3}+2x^{2}.
    (x^{3}+0x^{2}-5x-2)-(x^{3}+2x^{2})=-2x^{2}-5x-2
  5. -2x^{2}\div x=-2x; subtract -2x(x+2)=-2x^{2}-4x.
    (-2x^{2}-5x-2)-(-2x^{2}-4x)=-x-2
  6. -x\div x=-1; subtract -1(x+2)=-x-2.
    (-x-2)-(-x-2)=0
  7. So the quotient is x^{2}-2x-1.
    x^{3}-5x-2=(x+2)(x^{2}-2x-1)

The quadratic x^{2}-2x-1 does not factorise over the integers (its discriminant is 4+4=8, not a perfect square), so the factorisation stops here.

x^{3}-5x-2=(x+2)(x^{2}-2x-1)
Worked example: factorising when the first trial values fail
Factorise f(t)=t^{3}-3t^{2}-10t+24 by the Factor Theorem.
  1. A linear factor t-c needs c to divide the last term 24. Try c=1 first.
    f(1)=1-3-10+24=12\neq0
  2. Not zero, so t-1 is not a factor. Try c=-1.
    f(-1)=-1-3+10+24=30\neq0
  3. Try c=2.
    f(2)=8-12-20+24=0
  4. So t-2 is a factor. Divide: t^{3}\div t=t^{2}, then -t^{2}\div t=-t, then -12t\div t=-12, remainder 0.
    f(t)=(t-2)(t^{2}-t-12)
  5. Split the middle term of the quadratic: numbers with product -12 and sum -1 are -4 and 3.
    t^{2}-t-12=(t-4)(t+3)
  6. Combine the factors. Check the last term: (-2)(-4)(3)=24, as the Factor Theorem notes require.
    f(t)=(t-2)(t-4)(t+3)
The graph of f(t)=t^{3}-3t^{2}-10t+24 cuts the horizontal axis at -3, 2 and 4, matching the factors t+3, t-2, t-4.
(t-2)(t-4)(t+3)

5.6Applications of remainder theorem

If you give 10 pencils to five students out of 11, each student gets 2 pencils. Only one pencil remains with you, and this leftover pencil is called the remainder. Here 11 is the dividend, 5 is the divisor, 2 is the quotient and 1 is the remainder.

The remainder theorem is a powerful tool for solving a variety of mathematical problems. Here is a ball-height example from the book. Suppose Nasir hits a high fast ball straight up over home plate. The height of the ball after t seconds is

h(t)=-16t^{2}+80t+5

The roots of the function tell us at what times the ball is theoretically on the ground. When t=0 the height of the ball is 5 feet: this is the point at which he hits the ball. To find the height after 4 seconds, replace t by 4.

h(4)=-16(4)^{2}+80(4)+5=-256+320+5=69

After 4 seconds the height of the ball is 69 feet. Notice that h(4) is the same as the remainder when the polynomial is divided by t-4 (the Remainder Theorem).

Height of the ball in feet against time in seconds: it starts at 5 ft, is at 69 ft after 4 s, and returns to the ground a little after 5 s.
Example 11
The volume of a rectangular solid is 72 cubic units. The width is twice the height and the length is 7 units more than the height. Find the dimensions of the solid.
The solid of Example 11: height x, width 2x, length x+7.
  1. Let x be the height. Then the width is 2x and the length is x+7. Volume = width \times length \times height.
    \text{volume}=(2x)(x+7)(x)
  2. Put the volume equal to 72 and expand.
    72=2x^{3}+14x^{2}
  3. Divide by 2 and bring every term to the left side.
    x^{3}+7x^{2}-36=0
  4. Trace the possible zeros: try numbers that divide 36. For x=2:
    (2)^{3}+7(2)^{2}-36=8+28-36=0
  5. So 2 is a zero, that is, x-2 is a factor (Factor Theorem). Check that the other factor gives no positive root: dividing gives x^{3}+7x^{2}-36=(x-2)(x^{2}+9x+18)=(x-2)(x+3)(x+6), whose other zeros -3 and -6 are negative and cannot be lengths.
  6. Read the dimensions.
    \text{height}=x=2,\quad\text{width}=2x=4,\quad\text{length}=x+7=9
2\ \text{units},\ 4\ \text{units},\ 9\ \text{units}
Worked example: box dimensions from a volume polynomial
The volume of a rectangular box is V(x)=x^{3}+6x^{2}+11x+6 cubic units and its height is x+1. Find the other two sides.
The box has height x+1; the length and width are found by dividing the volume by the height.
  1. Volume is length times width times height, so the product of the other two sides is V(x)\div(x+1). First check that x+1 divides V exactly: put x=-1.
    V(-1)=-1+6-11+6=0
  2. The remainder is 0, so divide: x^{3}\div x=x^{2}, then 5x^{2}\div x=5x, then 6x\div x=6, remainder 0.
    V(x)=(x+1)(x^{2}+5x+6)
  3. Factorise the quadratic: numbers with product 6 and sum 5 are 2 and 3.
    x^{2}+5x+6=(x+2)(x+3)
  4. The other two sides are x+2 and x+3. Check with x=3: sides 4,5,6 give 120, and V(3)=27+54+33+6=120.
    V(x)=(x+1)(x+2)(x+3)
\text{sides }x+2\text{ and }x+3

Formula sheet

General form of a polynomial
P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\dots+a_{0}
a_n,\dots,a_0: coefficients; n: a whole number, the degree when a_n\neq0
Rational expression
\dfrac{P(x)}{Q(x)},\qquad Q(x)\neq0
P(x),Q(x): polynomials; Q is not the zero polynomial
Division relation
\text{dividend}=\text{quotient}\times\text{divisor}+\text{remainder}
degree of remainder is less than degree of divisor
Remainder theorem
p(x)=(x-c)\,q(x)+r\;\Longrightarrow\;r=p(c)
p(x): polynomial; c: constant; q(x): quotient; r: remainder on dividing by x-c
Zero of a polynomial
a\text{ is a zero of }P(x)\iff P(a)=0
a: a number; zeros are the roots of P(x)=0
Factor theorem
(x-c)\text{ is a factor of }p(x)\iff p(c)=0
c: a constant; if p(c)\neq0 then p(c) is the remainder
Factors of a cubic
f(x)=(x+a)(x+b)(x+c),\qquad abc=\text{last term of }f
a,b,c: positive or negative numbers

Key terms

Algebraic expression
A statement in which variables or constants or both are connected by the operations +,-,\times,\div.
Polynomial
An algebraic expression of one or more terms in which the exponents of the variables are whole numbers.
Degree of a polynomial
The highest exponent of the variable; with several variables, the largest sum of the exponents in a term.
Zero polynomial
The polynomial 0, whose coefficients are all zero; it has no degree.
Constant, linear, quadratic, cubic, bi-quadratic polynomial
Polynomials of degree 0, 1, 2, 3 and 4 respectively.
Rational expression
An expression \dfrac{P(x)}{Q(x)} with P, Q polynomials and Q not the zero polynomial.
Irrational expression
An algebraic expression that cannot be written as \dfrac{P(x)}{Q(x)} with P, Q polynomials.
Dividend, divisor, quotient, remainder
In p(x)=d(x)\,q(x)+r(x): p is the dividend, d the divisor, q the quotient and r the remainder, of lower degree than d.
Remainder theorem
If p(x) is divided by x-c, the remainder is p(c).
Zero of a polynomial
A number a with P(a)=0; the roots of the equation P(x)=0.
Factor theorem
x-c is a factor of p(x) if and only if p(c)=0.
Factor and divisor
Every factor is a divisor, but a divisor is a factor only when it divides the polynomial completely (remainder zero).

Exercise 5.1p. 156

The Remainder Theorem says the remainder on dividing p(x) by x-c is p(c); the Factor Theorem says x-c is a factor exactly when p(c)=0. Every answer below was checked against the answer key of the book.

1
Find the remainder of the following by using the Remainder Theorem when
Solution
(i) 2x^{3}+3x^{2}-4x+1 is divided by x+2.
  1. Let p(x)=2x^{3}+3x^{2}-4x+1. The divisor x+2=x-(-2) is zero at x=-2, so by the Remainder Theorem the remainder is p(-2).
    \text{Remainder}=p(-2)
  2. Substitute x=-2.
    p(-2)=2(-2)^{3}+3(-2)^{2}-4(-2)+1
  3. Evaluate the powers and multiply.
    =2(-8)+3(4)+8+1=-16+12+8+1
  4. Add.
    =5
5
(ii) x^{4}+2x^{3}-x^{2}+2x+3 is divided by x-2.
  1. Let p(x)=x^{4}+2x^{3}-x^{2}+2x+3. The divisor x-2 is zero at x=2, so the remainder is p(2).
    \text{Remainder}=p(2)
  2. Substitute x=2.
    p(2)=(2)^{4}+2(2)^{3}-(2)^{2}+2(2)+3
  3. Evaluate.
    =16+16-4+4+3=35
35

Book answer. (i) 5 (ii) 35

2
Show that x-3 is a factor of x^{3}-2x^{2}-5x+6.
Solution
  1. Let f(x)=x^{3}-2x^{2}-5x+6. By the Factor Theorem, x-3 is a factor if f(3)=0.
    f(3)=(3)^{3}-2(3)^{2}-5(3)+6
  2. Evaluate.
    =27-18-15+6=33-33=0
  3. Since f(3)=0, the remainder on dividing by x-3 is zero, so x-3 is a factor of f(x).
f(3)=0\ \Rightarrow\ x-3\text{ is a factor}
3
Decide whether x-3 is a factor of x^{3}-2x^{2}-5x+1 or not.
Solution
  1. Let f(x)=x^{3}-2x^{2}-5x+1. x-3 is a factor only if f(3)=0.
    f(3)=(3)^{3}-2(3)^{2}-5(3)+1
  2. Evaluate.
    =27-18-15+1=-5
  3. f(3)=-5\neq0, so the remainder is -5, not zero. Hence x-3 is not a factor.
f(3)=-5\neq0\ \Rightarrow\ x-3\text{ is not a factor}

Book answer. No

4
If 4y^{3}-4y^{2}+10+2y is completely divisible by any of its factor such that the quotient is 4y^{2}-8y+10, then find the other factor.
Solution

Arrange the polynomial in descending powers: 4y^{3}-4y^{2}+2y+10. Because it is divisible completely, dividend = factor \times quotient, so the other factor is the dividend divided by the given quotient.

  1. Divide 4y^{3}-4y^{2}+2y+10 by 4y^{2}-8y+10. Leading terms: 4y^{3}\div4y^{2}=y. Multiply and subtract.
    \begin{aligned}y(4y^{2}-8y+10)&=4y^{3}-8y^{2}+10y\\(4y^{3}-4y^{2}+2y+10)-(4y^{3}-8y^{2}+10y)&=4y^{2}-8y+10\end{aligned}
  2. Next: 4y^{2}\div4y^{2}=1. Multiply and subtract.
    (4y^{2}-8y+10)-1\cdot(4y^{2}-8y+10)=0
  3. The remainder is 0 and the quotient is y+1, so the other factor is y+1.
    4y^{3}-4y^{2}+2y+10=(y+1)(4y^{2}-8y+10)
  4. Check by the Factor Theorem: y=-1 must be a zero.
    4(-1)^{3}-4(-1)^{2}+2(-1)+10=-4-4-2+10=0
y+1

Book answer. y+1

5
Find the value of 'q' if x^{3}+qx^{2}-7x+6 is exactly divisible by (x+1).
Solution
  1. Let f(x)=x^{3}+qx^{2}-7x+6. Exactly divisible by x+1 means the remainder is zero, so f(-1)=0 (Factor Theorem).
    f(-1)=0
  2. Substitute x=-1.
    (-1)^{3}+q(-1)^{2}-7(-1)+6=0
  3. Simplify.
    -1+q+7+6=0\;\Longrightarrow\;q+12=0
  4. Solve for q.
    q=-12
q=-12

Book answer. -12

6
Find the value of 'm' in the polynomial 2x^{3}+3x^{2}-3x-m which when divided by x-2 gives the remainder of 16.
Solution
  1. Let p(x)=2x^{3}+3x^{2}-3x-m. By the Remainder Theorem the remainder on dividing by x-2 is p(2), and it is given to be 16.
    p(2)=16
  2. Substitute x=2.
    2(2)^{3}+3(2)^{2}-3(2)-m=16
  3. Simplify.
    16+12-6-m=16\;\Longrightarrow\;22-m=16
  4. Solve for m.
    m=6
m=6

Book answer. m=6

7
Check whether 1 and -2 are the zeros of x^{3}-7x+6.
Solution
  1. Let P(x)=x^{3}-7x+6. A number a is a zero of P(x) if and only if P(a)=0. Test a=1.
    P(1)=(1)^{3}-7(1)+6=1-7+6=0
  2. P(1)=0, so 1 is a zero. Now test a=-2.
    P(-2)=(-2)^{3}-7(-2)+6=-8+14+6=12
  3. P(-2)=12\neq0, so -2 is not a zero.
The curve meets the x-axis at x=1, 2 and -3 only. At x=-2 the height is 12, not 0 (hollow dot), so -2 is not a zero.
1\text{ is a zero; }-2\text{ is not}

Book answer. Only 1 is a zero of P(x)

8
Find zeros of the polynomial 2x^{3}+3x^{2}-11x-6.
Solution

The zeros are the values of x with p(x)=0. For a cubic, find one zero by trial, divide it out, and solve the quadratic that is left.

  1. Let p(x)=2x^{3}+3x^{2}-11x-6. Any zero that is a whole number divides the last term 6, so try \pm1,\pm2,\pm3,\pm6. Try x=2.
    p(2)=2(8)+3(4)-11(2)-6=16+12-22-6=0
  2. p(2)=0, so x-2 is a factor. Divide p(x) by x-2 to get the other factor.
  1. Divide the leading terms: 2x^{3}\div x=2x^{2}. Multiply the divisor by 2x^{2} and subtract the product.
    \begin{aligned}2x^{2}(x-2)&=2x^{3}-4x^{2}\\(2x^{3}+3x^{2}-11x-6)-(2x^{3}-4x^{2})&=7x^{2}-11x-6\end{aligned}
  2. Divide the leading terms: 7x^{2}\div x=7x. Multiply the divisor by 7x and subtract the product.
    \begin{aligned}7x(x-2)&=7x^{2}-14x\\(7x^{2}-11x-6)-(7x^{2}-14x)&=3x-6\end{aligned}
  3. Divide the leading terms: 3x\div x=3. Multiply the divisor by 3 and subtract the product.
    \begin{aligned}3(x-2)&=3x-6\\(3x-6)-(3x-6)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=2x^{2}+7x+3,\qquad\text{remainder}=0
  1. Factorise the quadratic quotient: split 7x as 6x+x.
    2x^{2}+7x+3=2x^{2}+6x+x+3=2x(x+3)+1(x+3)=(x+3)(2x+1)
  2. So the full factorisation is
    p(x)=(x-2)(x+3)(2x+1)
  3. Set each factor equal to zero.
    x-2=0\Rightarrow x=2;\quad x+3=0\Rightarrow x=-3;\quad 2x+1=0\Rightarrow x=-\dfrac12
The three zeros x=-3, -\tfrac12 and 2 are where the curve crosses the x-axis.
x=2,\ -3,\ -\dfrac12

Book answer. 2,\ -3,\ \dfrac{-1}{2}

9
Express f(x)=x^{3}-x^{2}-14x+8 in the form f(x)=(x-a)\,q(x)+r, where a=4.
Solution

The form f(x)=(x-a)q(x)+r is the division rule: dividend = divisor \times quotient + remainder, with divisor x-4. So divide f(x) by x-4.

  1. Divide the leading terms: x^{3}\div x=x^{2}. Multiply the divisor by x^{2} and subtract the product.
    \begin{aligned}x^{2}(x-4)&=x^{3}-4x^{2}\\(x^{3}-x^{2}-14x+8)-(x^{3}-4x^{2})&=3x^{2}-14x+8\end{aligned}
  2. Divide the leading terms: 3x^{2}\div x=3x. Multiply the divisor by 3x and subtract the product.
    \begin{aligned}3x(x-4)&=3x^{2}-12x\\(3x^{2}-14x+8)-(3x^{2}-12x)&=-2x+8\end{aligned}
  3. Divide the leading terms: -2x\div x=-2. Multiply the divisor by -2 and subtract the product.
    \begin{aligned}(-2)(x-4)&=-2x+8\\(-2x+8)-(-2x+8)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=x^{2}+3x-2,\qquad\text{remainder}=0
  1. Cross-check the remainder with the Remainder Theorem.
    r=f(4)=64-16-56+8=0
  2. Write the result in the required form.
    f(x)=(x-4)(x^{2}+3x-2)+0
f(x)=(x-4)(x^{2}+3x-2)+0

Book answer. f(x)=(x-4)(x^{2}+3x-2)+0

10
A rectangular room has a volume of (x^{3}+11x^{2}+34x+24) cubic feet. The height of the room is (x+1) feet. Find the area of its floor. (Hint: Volume of room = area of floor \times height.)
Solution
  1. By the hint, area of floor = volume \div height. Divide x^{3}+11x^{2}+34x+24 by x+1.
    \text{Area}=\dfrac{x^{3}+11x^{2}+34x+24}{x+1}
  1. Divide the leading terms: x^{3}\div x=x^{2}. Multiply the divisor by x^{2} and subtract the product.
    \begin{aligned}x^{2}(x+1)&=x^{3}+x^{2}\\(x^{3}+11x^{2}+34x+24)-(x^{3}+x^{2})&=10x^{2}+34x+24\end{aligned}
  2. Divide the leading terms: 10x^{2}\div x=10x. Multiply the divisor by 10x and subtract the product.
    \begin{aligned}10x(x+1)&=10x^{2}+10x\\(10x^{2}+34x+24)-(10x^{2}+10x)&=24x+24\end{aligned}
  3. Divide the leading terms: 24x\div x=24. Multiply the divisor by 24 and subtract the product.
    \begin{aligned}24(x+1)&=24x+24\\(24x+24)-(24x+24)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=x^{2}+10x+24,\qquad\text{remainder}=0
  1. The remainder is zero, so x+1 divides the volume exactly (check: p(-1)=-1+11-34+24=0), and the floor area is the quotient.
    \text{Area}=x^{2}+10x+24\ \text{square feet}
  2. The quotient factorises, which gives the floor sides.
    x^{2}+10x+24=(x+4)(x+6)
x^{2}+10x+24\ \text{ft}^{2}

Book answer. x^{2}+10x+24

Exercise 5.2p. 159

Factorise a cubic in three steps: find one zero c by trial (try numbers that divide the last term), divide by x-c, then factorise the quadratic quotient. A quadratic with a negative discriminant is left as it is.

1
Factorize y^{3}-7y-6 by using the factor theorem.
Solution
  1. Let f(y)=y^{3}-7y-6. Try numbers that divide the last term -6. Try y=-1.
    f(-1)=(-1)^{3}-7(-1)-6=-1+7-6=0
  2. f(-1)=0, so y+1 is a factor (Factor Theorem). Divide y^{3}+0y^{2}-7y-6 by y+1.
  1. Divide the leading terms: y^{3}\div y=y^{2}. Multiply the divisor by y^{2} and subtract the product.
    \begin{aligned}y^{2}(y+1)&=y^{3}+y^{2}\\(y^{3}-7y-6)-(y^{3}+y^{2})&=-y^{2}-7y-6\end{aligned}
  2. Divide the leading terms: -y^{2}\div y=-y. Multiply the divisor by -y and subtract the product.
    \begin{aligned}(-y)(y+1)&=-y^{2}-y\\(-y^{2}-7y-6)-(-y^{2}-y)&=-6y-6\end{aligned}
  3. Divide the leading terms: -6y\div y=-6. Multiply the divisor by -6 and subtract the product.
    \begin{aligned}(-6)(y+1)&=-6y-6\\(-6y-6)-(-6y-6)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=y^{2}-y-6,\qquad\text{remainder}=0
  1. Factorise the quadratic y^{2}-y-6: find two numbers with product -6 and sum -1, namely -3 and 2.
    y^{2}-y-6=y^{2}-3y+2y-6=y(y-3)+2(y-3)=(y-3)(y+2)
  2. Collect all factors.
    y^{3}-7y-6=(y+1)(y-3)(y+2)
  3. Check with the last-term rule.
    (+1)(-3)(+2)=-6
(y+1)(y-3)(y+2)

Book answer. (y+1)(y-3)(y+2)

2
Factorize 2x^{3}-x^{2}-2x+1 by using the factor theorem.
Solution
  1. Let f(x)=2x^{3}-x^{2}-2x+1. Try x=1.
    f(1)=2-1-2+1=0
  2. f(1)=0, so x-1 is a factor. Divide f(x) by x-1.
  1. Divide the leading terms: 2x^{3}\div x=2x^{2}. Multiply the divisor by 2x^{2} and subtract the product.
    \begin{aligned}2x^{2}(x-1)&=2x^{3}-2x^{2}\\(2x^{3}-x^{2}-2x+1)-(2x^{3}-2x^{2})&=x^{2}-2x+1\end{aligned}
  2. Divide the leading terms: x^{2}\div x=x. Multiply the divisor by x and subtract the product.
    \begin{aligned}x(x-1)&=x^{2}-x\\(x^{2}-2x+1)-(x^{2}-x)&=-x+1\end{aligned}
  3. Divide the leading terms: -x\div x=-1. Multiply the divisor by -1 and subtract the product.
    \begin{aligned}(-1)(x-1)&=-x+1\\(-x+1)-(-x+1)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=2x^{2}+x-1,\qquad\text{remainder}=0
  1. Factorise 2x^{2}+x-1: product 2\times(-1)=-2, sum 1, so split x=2x-x.
    2x^{2}+x-1=2x^{2}+2x-x-1=2x(x+1)-1(x+1)=(x+1)(2x-1)
  2. Collect all factors.
    2x^{3}-x^{2}-2x+1=(x-1)(x+1)(2x-1)
(x-1)(x+1)(2x-1)

Book answer. (x-1)(x+1)(2x-1)

3
Factorize 2x^{3}+5x^{2}-9x-18 by using the factor theorem.
Solution
  1. Let f(x)=2x^{3}+5x^{2}-9x-18. Try x=2.
    f(2)=2(8)+5(4)-9(2)-18=16+20-18-18=0
  2. f(2)=0, so x-2 is a factor. Divide f(x) by x-2.
  1. Divide the leading terms: 2x^{3}\div x=2x^{2}. Multiply the divisor by 2x^{2} and subtract the product.
    \begin{aligned}2x^{2}(x-2)&=2x^{3}-4x^{2}\\(2x^{3}+5x^{2}-9x-18)-(2x^{3}-4x^{2})&=9x^{2}-9x-18\end{aligned}
  2. Divide the leading terms: 9x^{2}\div x=9x. Multiply the divisor by 9x and subtract the product.
    \begin{aligned}9x(x-2)&=9x^{2}-18x\\(9x^{2}-9x-18)-(9x^{2}-18x)&=9x-18\end{aligned}
  3. Divide the leading terms: 9x\div x=9. Multiply the divisor by 9 and subtract the product.
    \begin{aligned}9(x-2)&=9x-18\\(9x-18)-(9x-18)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=2x^{2}+9x+9,\qquad\text{remainder}=0
  1. Factorise 2x^{2}+9x+9: product 2\times9=18, sum 9, so split 9x=6x+3x.
    2x^{2}+9x+9=2x^{2}+6x+3x+9=2x(x+3)+3(x+3)=(x+3)(2x+3)
  2. Collect all factors.
    2x^{3}+5x^{2}-9x-18=(x-2)(x+3)(2x+3)
(x-2)(x+3)(2x+3)

Book answer. (x-2)(x+3)(2x+3)

4
Factorize 3x^{3}-5x^{2}-36 by using the factor theorem.
Solution
  1. Let f(x)=3x^{3}-5x^{2}+0x-36 (there is no x term). Try numbers that divide 36. Try x=3.
    f(3)=3(27)-5(9)-36=81-45-36=0
  2. f(3)=0, so x-3 is a factor. Divide f(x) by x-3.
  1. Divide the leading terms: 3x^{3}\div x=3x^{2}. Multiply the divisor by 3x^{2} and subtract the product.
    \begin{aligned}3x^{2}(x-3)&=3x^{3}-9x^{2}\\(3x^{3}-5x^{2}-36)-(3x^{3}-9x^{2})&=4x^{2}-36\end{aligned}
  2. Divide the leading terms: 4x^{2}\div x=4x. Multiply the divisor by 4x and subtract the product.
    \begin{aligned}4x(x-3)&=4x^{2}-12x\\(4x^{2}-36)-(4x^{2}-12x)&=12x-36\end{aligned}
  3. Divide the leading terms: 12x\div x=12. Multiply the divisor by 12 and subtract the product.
    \begin{aligned}12(x-3)&=12x-36\\(12x-36)-(12x-36)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=3x^{2}+4x+12,\qquad\text{remainder}=0
  1. Test whether 3x^{2}+4x+12 factorises: its discriminant is
    b^{2}-4ac=16-4(3)(12)=16-144=-128<0
  2. The discriminant is negative, so the quadratic has no real factors. The factorisation is complete.
    3x^{3}-5x^{2}-36=(x-3)(3x^{2}+4x+12)
(x-3)(3x^{2}+4x+12)

Book answer. (x-3)(3x^{2}+4x+12)

5
Factorize t^{3}+t^{2}+3t-5 by using the factor theorem.
Solution
  1. Let f(t)=t^{3}+t^{2}+3t-5. Try numbers that divide 5. Try t=1.
    f(1)=1+1+3-5=0
  2. f(1)=0, so t-1 is a factor. Divide f(t) by t-1.
  1. Divide the leading terms: t^{3}\div t=t^{2}. Multiply the divisor by t^{2} and subtract the product.
    \begin{aligned}t^{2}(t-1)&=t^{3}-t^{2}\\(t^{3}+t^{2}+3t-5)-(t^{3}-t^{2})&=2t^{2}+3t-5\end{aligned}
  2. Divide the leading terms: 2t^{2}\div t=2t. Multiply the divisor by 2t and subtract the product.
    \begin{aligned}2t(t-1)&=2t^{2}-2t\\(2t^{2}+3t-5)-(2t^{2}-2t)&=5t-5\end{aligned}
  3. Divide the leading terms: 5t\div t=5. Multiply the divisor by 5 and subtract the product.
    \begin{aligned}5(t-1)&=5t-5\\(5t-5)-(5t-5)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=t^{2}+2t+5,\qquad\text{remainder}=0
  1. The discriminant of t^{2}+2t+5 is
    b^{2}-4ac=4-20=-16<0
  2. It is negative, so the quadratic does not factorise over the real numbers.
    t^{3}+t^{2}+3t-5=(t-1)(t^{2}+2t+5)
(t-1)(t^{2}+2t+5)

Book answer. (t-1)(t^{2}+2t+5)

62026 Annual I, Set B, Q.6 (b)
If (x-2) is one of the factor of 2x^{3}-15x^{2}+16x+12, find its other factors.
Solution
  1. Let f(x)=2x^{3}-15x^{2}+16x+12. Check the given factor with the Factor Theorem: f(2) should be 0.
    f(2)=2(8)-15(4)+16(2)+12=16-60+32+12=0
  2. Divide f(x) by x-2 to get the remaining quadratic factor.
  1. Divide the leading terms: 2x^{3}\div x=2x^{2}. Multiply the divisor by 2x^{2} and subtract the product.
    \begin{aligned}2x^{2}(x-2)&=2x^{3}-4x^{2}\\(2x^{3}-15x^{2}+16x+12)-(2x^{3}-4x^{2})&=-11x^{2}+16x+12\end{aligned}
  2. Divide the leading terms: -11x^{2}\div x=-11x. Multiply the divisor by -11x and subtract the product.
    \begin{aligned}(-11x)(x-2)&=-11x^{2}+22x\\(-11x^{2}+16x+12)-(-11x^{2}+22x)&=-6x+12\end{aligned}
  3. Divide the leading terms: -6x\div x=-6. Multiply the divisor by -6 and subtract the product.
    \begin{aligned}(-6)(x-2)&=-6x+12\\(-6x+12)-(-6x+12)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=2x^{2}-11x-6,\qquad\text{remainder}=0
  1. Factorise 2x^{2}-11x-6: product 2\times(-6)=-12, sum -11, so split -11x=-12x+x.
    2x^{2}-11x-6=2x^{2}-12x+x-6=2x(x-6)+1(x-6)=(x-6)(2x+1)
  2. So f(x)=(x-2)(x-6)(2x+1) and the other two factors are x-6 and 2x+1.
(x-6)\ \text{and}\ (2x+1)

Book answer. Other two factors are (x-6) and (2x+1)

7
Factorize 2x^{3}-15x^{2}+27x-10 if \dfrac12 is one of its zero.
Solution
  1. If x=\tfrac12 is a zero, then x-\tfrac12=0, so 2x-1 is a factor (Factor Theorem). Check:
    f\left(\tfrac12\right)=2\cdot\tfrac18-15\cdot\tfrac14+27\cdot\tfrac12-10=\tfrac14-\tfrac{15}{4}+\tfrac{27}{2}-10=0
  2. Divide 2x^{3}-15x^{2}+27x-10 by 2x-1.
  1. Divide the leading terms: 2x^{3}\div 2x=x^{2}. Multiply the divisor by x^{2} and subtract the product.
    \begin{aligned}x^{2}(2x-1)&=2x^{3}-x^{2}\\(2x^{3}-15x^{2}+27x-10)-(2x^{3}-x^{2})&=-14x^{2}+27x-10\end{aligned}
  2. Divide the leading terms: -14x^{2}\div 2x=-7x. Multiply the divisor by -7x and subtract the product.
    \begin{aligned}(-7x)(2x-1)&=-14x^{2}+7x\\(-14x^{2}+27x-10)-(-14x^{2}+7x)&=20x-10\end{aligned}
  3. Divide the leading terms: 20x\div 2x=10. Multiply the divisor by 10 and subtract the product.
    \begin{aligned}10(2x-1)&=20x-10\\(20x-10)-(20x-10)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=x^{2}-7x+10,\qquad\text{remainder}=0
  1. Factorise x^{2}-7x+10: two numbers with product 10 and sum -7 are -2 and -5.
    x^{2}-7x+10=(x-2)(x-5)
  2. Collect all factors.
    2x^{3}-15x^{2}+27x-10=(2x-1)(x-2)(x-5)
(2x-1)(x-2)(x-5)

Book answer. (2x-1)(x-5)(x-2)

8
If h(x)=4x^{3}+4x^{2}+73x+36 and h\left(\dfrac{-1}{2}\right)=0, then factorize h(x).
Solution
  1. Verify the given value.
    h\left(-\tfrac12\right)=4\left(-\tfrac18\right)+4\left(\tfrac14\right)+73\left(-\tfrac12\right)+36=-\tfrac12+1-\tfrac{73}{2}+36=0
  2. h(-\tfrac12)=0, so x+\tfrac12, that is 2x+1, is a factor (Factor Theorem). Divide h(x) by 2x+1.
  1. Divide the leading terms: 4x^{3}\div 2x=2x^{2}. Multiply the divisor by 2x^{2} and subtract the product.
    \begin{aligned}2x^{2}(2x+1)&=4x^{3}+2x^{2}\\(4x^{3}+4x^{2}+73x+36)-(4x^{3}+2x^{2})&=2x^{2}+73x+36\end{aligned}
  2. Divide the leading terms: 2x^{2}\div 2x=x. Multiply the divisor by x and subtract the product.
    \begin{aligned}x(2x+1)&=2x^{2}+x\\(2x^{2}+73x+36)-(2x^{2}+x)&=72x+36\end{aligned}
  3. Divide the leading terms: 72x\div 2x=36. Multiply the divisor by 36 and subtract the product.
    \begin{aligned}36(2x+1)&=72x+36\\(72x+36)-(72x+36)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=2x^{2}+x+36,\qquad\text{remainder}=0
  1. Check whether 2x^{2}+x+36 factorises: its discriminant is
    b^{2}-4ac=1-4(2)(36)=1-288=-287<0
  2. It is negative, so there are no more real factors.
    h(x)=(2x+1)(2x^{2}+x+36)
(2x+1)(2x^{2}+x+36)

Book answer. (2x+1)(2x^{2}+x+36)

Exercise 5.3p. 160

Turn the words into a cubic equation in one unknown (volume = length \times width \times height), find a positive zero by trial with the Factor Theorem, and read the dimensions from it.

1
The volume of a wooden box is 120 cubic centimeters. The length of the box is 7 cm longer than its height, and the width is 3 cm shorter than its height. Find the dimensions of the box.
Solution
  1. Let the height be h cm. Then length =h+7 and width =h-3.
    \text{height}=h,\quad\text{length}=h+7,\quad\text{width}=h-3
  2. Volume = length \times width \times height =120.
    h(h+7)(h-3)=120
  3. Multiply (h+7)(h-3)=h^{2}+4h-21, then by h, and bring 120 to the left.
    h^{3}+4h^{2}-21h-120=0
  4. Trace a zero: try numbers that divide 120. For h=5:
    (5)^{3}+4(5)^{2}-21(5)-120=125+100-105-120=0
  5. So h-5 is a factor. Divide the cubic by h-5.
  1. Divide the leading terms: h^{3}\div h=h^{2}. Multiply the divisor by h^{2} and subtract the product.
    \begin{aligned}h^{2}(h-5)&=h^{3}-5h^{2}\\(h^{3}+4h^{2}-21h-120)-(h^{3}-5h^{2})&=9h^{2}-21h-120\end{aligned}
  2. Divide the leading terms: 9h^{2}\div h=9h. Multiply the divisor by 9h and subtract the product.
    \begin{aligned}9h(h-5)&=9h^{2}-45h\\(9h^{2}-21h-120)-(9h^{2}-45h)&=24h-120\end{aligned}
  3. Divide the leading terms: 24h\div h=24. Multiply the divisor by 24 and subtract the product.
    \begin{aligned}24(h-5)&=24h-120\\(24h-120)-(24h-120)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=h^{2}+9h+24,\qquad\text{remainder}=0
  1. The quadratic h^{2}+9h+24 has discriminant 81-96=-15<0, so it has no real zeros. The only real solution is h=5.
    h^{3}+4h^{2}-21h-120=(h-5)(h^{2}+9h+24)
  2. Read off the dimensions.
    \text{height}=5\ \text{cm},\quad\text{length}=5+7=12\ \text{cm},\quad\text{width}=5-3=2\ \text{cm}
  3. Check: 5\times12\times2=120.
5\ \text{cm}\times12\ \text{cm}\times2\ \text{cm}

Book answer. 5 cm by 12 cm by 2 cm

2
In the cricket match season, the number of tickets sold during the match can be modeled by t(x)=x^{3}-12x^{2}+48x+74, where x is the number of games played. Find the number of tickets sold during the twelfth game of the cricket season.
Solution
  1. The twelfth game means x=12. The number of tickets is t(12), which is also the remainder when t(x) is divided by x-12 (Remainder Theorem).
    t(12)=(12)^{3}-12(12)^{2}+48(12)+74
  2. Evaluate each term.
    =1728-1728+576+74
  3. Add. The first two terms cancel.
    =576+74=650
650\ \text{tickets}

Book answer. 650

3
A rectangular solid has a volume of 144 cubic units. The width is twice the height and the length is 2 units more than the width. Find the dimensions of the solid.
Solution
  1. Let the height be h. Then width =2h and length =2h+2.
    \text{height}=h,\quad\text{width}=2h,\quad\text{length}=2h+2
  2. Volume =144.
    h\cdot2h\cdot(2h+2)=144
  3. Expand: 2h^{2}(2h+2)=4h^{3}+4h^{2}. Divide by 4 and bring 36 to the left.
    4h^{3}+4h^{2}=144\;\Longrightarrow\;h^{3}+h^{2}-36=0
  4. Trace a zero: try numbers that divide 36. For h=3:
    (3)^{3}+(3)^{2}-36=27+9-36=0
  5. So h-3 is a factor. Divide h^{3}+h^{2}+0h-36 by h-3.
  1. Divide the leading terms: h^{3}\div h=h^{2}. Multiply the divisor by h^{2} and subtract the product.
    \begin{aligned}h^{2}(h-3)&=h^{3}-3h^{2}\\(h^{3}+h^{2}-36)-(h^{3}-3h^{2})&=4h^{2}-36\end{aligned}
  2. Divide the leading terms: 4h^{2}\div h=4h. Multiply the divisor by 4h and subtract the product.
    \begin{aligned}4h(h-3)&=4h^{2}-12h\\(4h^{2}-36)-(4h^{2}-12h)&=12h-36\end{aligned}
  3. Divide the leading terms: 12h\div h=12. Multiply the divisor by 12 and subtract the product.
    \begin{aligned}12(h-3)&=12h-36\\(12h-36)-(12h-36)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=h^{2}+4h+12,\qquad\text{remainder}=0
  1. The quadratic h^{2}+4h+12 has discriminant 16-48=-32<0, so h=3 is the only real zero.
  2. Read off the dimensions.
    \text{height}=3,\quad\text{width}=2(3)=6,\quad\text{length}=2(3)+2=8
  3. Check: 3\times6\times8=144.
3\ \text{units},\ 6\ \text{units},\ 8\ \text{units}

Book answer. 6 units by 8 units by 3 units

4
The volume of a rectangular solid is 2475 cubic units. The length of the box is three units more than twice the width of the box. The height is 2 units less than width. Find the dimensions of the box.
Solution
  1. Let the width be w. Then length =2w+3 and height =w-2.
    \text{width}=w,\quad\text{length}=2w+3,\quad\text{height}=w-2
  2. Volume =2475.
    w(2w+3)(w-2)=2475
  3. Expand: (2w+3)(w-2)=2w^{2}-w-6, so w(2w^{2}-w-6)=2w^{3}-w^{2}-6w.
    2w^{3}-w^{2}-6w-2475=0
  4. Trace a zero. The height w-2 must be positive, so w>2. Try numbers that divide 2475. For w=9 the value is negative, for w=11 it is zero.
    \begin{aligned}w=9:&\ 2(729)-81-54-2475=-1152\\w=11:&\ 2(1331)-121-66-2475=2662-2662=0\end{aligned}
  5. So w-11 is a factor. Divide the cubic by w-11.
  1. Divide the leading terms: 2w^{3}\div w=2w^{2}. Multiply the divisor by 2w^{2} and subtract the product.
    \begin{aligned}2w^{2}(w-11)&=2w^{3}-22w^{2}\\(2w^{3}-w^{2}-6w-2475)-(2w^{3}-22w^{2})&=21w^{2}-6w-2475\end{aligned}
  2. Divide the leading terms: 21w^{2}\div w=21w. Multiply the divisor by 21w and subtract the product.
    \begin{aligned}21w(w-11)&=21w^{2}-231w\\(21w^{2}-6w-2475)-(21w^{2}-231w)&=225w-2475\end{aligned}
  3. Divide the leading terms: 225w\div w=225. Multiply the divisor by 225 and subtract the product.
    \begin{aligned}225(w-11)&=225w-2475\\(225w-2475)-(225w-2475)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=2w^{2}+21w+225,\qquad\text{remainder}=0
  1. The quadratic 2w^{2}+21w+225 has discriminant 441-1800<0, so w=11 is the only real zero.
  2. Read off the dimensions.
    \text{width}=11,\quad\text{length}=2(11)+3=25,\quad\text{height}=11-2=9
  3. Check: 11\times25\times9=2475.
9\ \text{units},\ 11\ \text{units},\ 25\ \text{units}

Book answer. 9 units by 11 units by 25 units

5
The area of rectangle ACED is represented by 6x^{2}+38x+56. Its width is represented by 2x+8. Point B is the midpoint of AC. ABFG is a square. Find the length of rectangle ACED and the area of square ABFG.
Solution
The rectangle ACED of Question 5. B is the midpoint of AC and ABFG is a square.
  1. Area of a rectangle = length \times width, so length = area \div width. Divide 6x^{2}+38x+56 by 2x+8.
    \text{Length}=AD=\dfrac{6x^{2}+38x+56}{2x+8}
  1. Divide the leading terms: 6x^{2}\div 2x=3x. Multiply the divisor by 3x and subtract the product.
    \begin{aligned}3x(2x+8)&=6x^{2}+24x\\(6x^{2}+38x+56)-(6x^{2}+24x)&=14x+56\end{aligned}
  2. Divide the leading terms: 14x\div 2x=7. Multiply the divisor by 7 and subtract the product.
    \begin{aligned}7(2x+8)&=14x+56\\(14x+56)-(14x+56)&=0\end{aligned}
  3. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=3x+7,\qquad\text{remainder}=0
  1. The remainder is zero, so the length of ACED is the quotient.
    AD=3x+7
  2. The width is AC=2x+8 and B is its midpoint, so AB=\dfrac{2x+8}{2}=x+4. ABFG is a square, so its side is x+4.
    AB=x+4
  3. Area of the square =(\text{side})^{2}.
    \text{Area of ABFG}=(x+4)^{2}=x^{2}+8x+16
\text{Length of ACED}=3x+7;\quad\text{side of ABFG}=x+4,\ \text{area}=(x+4)^{2}

Book answer. Length of one side of square ABFG is x+4. Area =(x+4)^{2}. The length of rectangle ACED =3x+7.

6
The volume of the box is y^{3}-2y^{2}-y+2. If the length of one side is y-2, find the length of the other two sides.
Solution

Volume = length \times width \times height, so the product of the other two sides is the volume divided by y-2.

  1. Divide the leading terms: y^{3}\div y=y^{2}. Multiply the divisor by y^{2} and subtract the product.
    \begin{aligned}y^{2}(y-2)&=y^{3}-2y^{2}\\(y^{3}-2y^{2}-y+2)-(y^{3}-2y^{2})&=-y+2\end{aligned}
  2. Divide the leading terms: -y\div y=-1. Multiply the divisor by -1 and subtract the product.
    \begin{aligned}(-1)(y-2)&=-y+2\\(-y+2)-(-y+2)&=0\end{aligned}
  3. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=y^{2}-1,\qquad\text{remainder}=0
  1. Factorise the quotient y^{2}-1 as a difference of squares. These are the other two sides.
    y^{2}-1=(y-1)(y+1)
  2. Check: (y-2)(y-1)(y+1)=(y-2)(y^{2}-1)=y^{3}-y-2y^{2}+2=y^{3}-2y^{2}-y+2.
y+1\ \text{and}\ y-1

Book answer. y+1,\ y-1

Review Exercise 5p. 161

A mixed review. The multiple-choice parts are answered with the Remainder and Factor Theorems; the other questions practise division, factors and zeros.

1 (i)
Factors of -2-x+x^{2} are:
  1. A(x-2)(x-1)
  2. B(x+1)(x+2)
  3. C(x+2)(x-1)
  4. D(x+1)(x-2)Answer
Why

Write in descending powers: x^{2}-x-2. Find two numbers with product -2 and sum -1: they are -2 and +1.

x^{2}-x-2=x^{2}-2x+x-2=x(x-2)+1(x-2)=(x+1)(x-2)

Option (a) has product +2 and option (c) has sum +1, so neither is correct.

1 (ii)
Divide 9y^{2}+9y-10 by 3y-2, then remainder is:
  1. A0Answer
  2. B1
  3. C2
  4. D3
Why

By the Remainder Theorem the remainder is the value at y=\tfrac23, where 3y-2=0.

9\left(\tfrac23\right)^{2}+9\left(\tfrac23\right)-10=4+6-10=0

So 3y-2 divides exactly (the quotient is 3y+5) and the remainder is 0.

1 (iii)
\dfrac{x^{2}-x-9}{x-3}=x+2+\dfrac{?}{x-3}
  1. A-27
  2. B-3Answer
  3. C\dfrac{3}{x-3}+x+2
  4. D3
Why

The numerator over the denominator is the quotient plus remainder over divisor. The remainder is p(3) with p(x)=x^{2}-x-9.

p(3)=9-3-9=-3

Check: (x+2)(x-3)=x^{2}-x-6, and x^{2}-x-9=(x^{2}-x-6)-3, so the missing numerator is -3.

1 (iv)
If 3x^{3}-2x^{2}+5 is divided by x+1, then x+1 will be its:
  1. Adivisor as well as factorAnswer
  2. Bdividend
  3. Cquotient
  4. Dremainder
Why

x+1 is the polynomial we divide by, so it is the divisor. Test whether it is also a factor: the remainder is the value at x=-1.

3(-1)^{3}-2(-1)^{2}+5=-3-2+5=0

The remainder is 0, so x+1 divides completely: it is a divisor as well as a factor. It is not the dividend, the quotient or the remainder.

1 (v)
If 2 is a zero of the polynomial x^{3}+5x^{2}-4x+k, then the value of k will be:
  1. A-4
  2. B-20Answer
  3. C20
  4. D0
Why

A zero a satisfies P(a)=0, so put x=2 equal to zero.

8+20-8+k=0\;\Longrightarrow\;20+k=0\;\Longrightarrow\;k=-20
1 (vi)
If x-b is the factor of q(x), then q(b) is:
  1. Afactor
  2. Bdivisor
  3. CremainderAnswer
  4. Ddividend
Why

By the Remainder Theorem q(b) is the remainder when q(x) is divided by x-b. Since x-b is a factor, this remainder is 0.

q(b) is a number, not a polynomial, so it cannot be the factor, the divisor (x-b is the divisor) or the dividend (q(x) is the dividend). It is the remainder.

1 (vii)
If the expression 2x^{3}+3px^{2}-4x has a remainder of 4 when divided by x+2, then p=
  1. A-2
  2. B1Answer
  3. C-1
  4. D0
Why

The remainder on dividing by x+2 is the value at x=-2, and it is given as 4.

2(-2)^{3}+3p(-2)^{2}-4(-2)=-16+12p+8=12p-8
12p-8=4\;\Longrightarrow\;12p=12\;\Longrightarrow\;p=1
1 (viii)
If f(x) is divided by x-2, then remainder is 12. What is f(2)?
  1. A-12
  2. Bf(-2)
  3. C12Answer
  4. Dzero
Why

By the Remainder Theorem the remainder on dividing by x-2 is f(2), so f(2)=12.

f(-2) would be the remainder for the divisor x+2, so option (b) is not correct.

2
(64y^{3}-8)\div(4y-2)
Solution

Insert the missing terms: 64y^{3}-8=64y^{3}+0y^{2}+0y-8, then use long division.

  1. Divide the leading terms: 64y^{3}\div 4y=16y^{2}. Multiply the divisor by 16y^{2} and subtract the product.
    \begin{aligned}16y^{2}(4y-2)&=64y^{3}-32y^{2}\\(64y^{3}-8)-(64y^{3}-32y^{2})&=32y^{2}-8\end{aligned}
  2. Divide the leading terms: 32y^{2}\div 4y=8y. Multiply the divisor by 8y and subtract the product.
    \begin{aligned}8y(4y-2)&=32y^{2}-16y\\(32y^{2}-8)-(32y^{2}-16y)&=16y-8\end{aligned}
  3. Divide the leading terms: 16y\div 4y=4. Multiply the divisor by 4 and subtract the product.
    \begin{aligned}4(4y-2)&=16y-8\\(16y-8)-(16y-8)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=16y^{2}+8y+4,\qquad\text{remainder}=0
  1. Check by multiplying the divisor by the quotient.
    (4y-2)(16y^{2}+8y+4)=64y^{3}+32y^{2}+16y-32y^{2}-16y-8=64y^{3}-8
  2. Another way: 64y^{3}-8=(4y)^{3}-2^{3} and a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2}) gives the same quotient (4y)^{2}+(4y)(2)+2^{2}.
\text{Quotient }16y^{2}+8y+4,\ \text{remainder }0

Book answer. 16y^{2}+4y+4

Note. The book's answer key prints 16y^{2}+4y+4. Multiplying back, (4y-2)(16y^{2}+4y+4)=64y^{3}-8y^{2}-\dots does not give 64y^{3}-8, so the correct quotient is 16y^{2}+8y+4.

3
(125y^{3}-8)\div(5y-2)
Solution

Insert the missing terms: 125y^{3}-8=125y^{3}+0y^{2}+0y-8.

  1. Divide the leading terms: 125y^{3}\div 5y=25y^{2}. Multiply the divisor by 25y^{2} and subtract the product.
    \begin{aligned}25y^{2}(5y-2)&=125y^{3}-50y^{2}\\(125y^{3}-8)-(125y^{3}-50y^{2})&=50y^{2}-8\end{aligned}
  2. Divide the leading terms: 50y^{2}\div 5y=10y. Multiply the divisor by 10y and subtract the product.
    \begin{aligned}10y(5y-2)&=50y^{2}-20y\\(50y^{2}-8)-(50y^{2}-20y)&=20y-8\end{aligned}
  3. Divide the leading terms: 20y\div 5y=4. Multiply the divisor by 4 and subtract the product.
    \begin{aligned}4(5y-2)&=20y-8\\(20y-8)-(20y-8)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=25y^{2}+10y+4,\qquad\text{remainder}=0
  1. Check with a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2}) for a=5y, b=2.
    (5y)^{3}-2^{3}=(5y-2)\big(25y^{2}+10y+4\big)
\text{Quotient }25y^{2}+10y+4,\ \text{remainder }0

Book answer. 25y^{2}+10y+4

4
Is 3y-2 a factor of 6y^{3}-y^{2}-5y+2?
Solution
  1. 3y-2 is zero at y=\tfrac23. By the Factor Theorem it is a factor if f\left(\tfrac23\right)=0, where f(y)=6y^{3}-y^{2}-5y+2.
    f\left(\tfrac23\right)=6\cdot\tfrac{8}{27}-\tfrac49-\tfrac{10}{3}+2
  2. Put everything over 9.
    =\tfrac{16}{9}-\tfrac49-\tfrac{30}{9}+\tfrac{18}{9}=\tfrac{16-4-30+18}{9}=0
  3. The remainder is zero, so 3y-2 is a factor.

Division confirms it with no remainder:

  1. Divide the leading terms: 6y^{3}\div 3y=2y^{2}. Multiply the divisor by 2y^{2} and subtract the product.
    \begin{aligned}2y^{2}(3y-2)&=6y^{3}-4y^{2}\\(6y^{3}-y^{2}-5y+2)-(6y^{3}-4y^{2})&=3y^{2}-5y+2\end{aligned}
  2. Divide the leading terms: 3y^{2}\div 3y=y. Multiply the divisor by y and subtract the product.
    \begin{aligned}y(3y-2)&=3y^{2}-2y\\(3y^{2}-5y+2)-(3y^{2}-2y)&=-3y+2\end{aligned}
  3. Divide the leading terms: -3y\div 3y=-1. Multiply the divisor by -1 and subtract the product.
    \begin{aligned}(-1)(3y-2)&=-3y+2\\(-3y+2)-(-3y+2)&=0\end{aligned}
  4. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=2y^{2}+y-1,\qquad\text{remainder}=0
\text{Yes, } 6y^{3}-y^{2}-5y+2=(3y-2)(2y^{2}+y-1)

Book answer. Yes

5
If zeros of a polynomial are 4,\ \dfrac35,\ -2, find the polynomial.
Solution
  1. If a is a zero then x-a is a factor. Write the polynomial as a product of the three factors. The factor for \tfrac35 is x-\tfrac35; multiplying it by 5 gives 5x-3, which has whole-number coefficients.
    (x-4)\left(x-\tfrac35\right)(x+2)\ \sim\ (x-4)(5x-3)(x+2)
  2. Multiply the factors with whole-number coefficients (x-4)(x+2) first.
    (x-4)(x+2)=x^{2}+2x-4x-8=x^{2}-2x-8
  3. Multiply by 5x-3.
    (x^{2}-2x-8)(5x-3)=5x^{3}-3x^{2}-10x^{2}+6x-40x+24
  4. Collect like terms.
    =5x^{3}-13x^{2}-34x+24
5x^{3}-13x^{2}-34x+24

Book answer. 5x^{3}-13x^{2}-34x+24

Note. Multiplying by 5 only scales the polynomial by a constant, so its zeros do not change.

6
Find the value of 'k' so that the remainder upon dividing (x^{2}+8x+k) by (x-4) is zero.
Solution
  1. By the Remainder Theorem the remainder on dividing by x-4 is the value at x=4. Set it equal to zero.
    (4)^{2}+8(4)+k=0
  2. Simplify and solve.
    16+32+k=0\;\Longrightarrow\;k=-48
  3. Check: x^{2}+8x-48=(x-4)(x+12).
k=-48

Book answer. -48

7
Suppose that the quotient upon dividing one polynomial by another is 3x^{2}-x+32-\dfrac{121}{x+4}. What is the dividend?
Solution

A quotient written as 3x^{2}-x+32-\dfrac{121}{x+4} means: quotient 3x^{2}-x+32, remainder -121, divisor x+4. Use dividend = quotient \times divisor + remainder.

  1. Substitute the three parts.
    \text{dividend}=(3x^{2}-x+32)(x+4)-121
  2. Expand the product.
    =3x^{3}+12x^{2}-x^{2}-4x+32x+128-121
  3. Collect like terms.
    =3x^{3}+11x^{2}+28x+7
  4. Check: dividing this by x+4 gives remainder 3(-64)+11(16)+28(-4)+7=-192+176-112+7=-121.
3x^{3}+11x^{2}+28x+7

Book answer. x+4

Note. The book's answer key prints x+4, which is the divisor, not the dividend. The dividend is 3x^{3}+11x^{2}+28x+7.

8
If two linear factors of the polynomial y^{3}+6y^{2}-y-30 are (y-2) and (y+3), find its third factor.
Solution
  1. The product of the two known factors is a quadratic factor of the cubic.
    (y-2)(y+3)=y^{2}+3y-2y-6=y^{2}+y-6
  2. Divide y^{3}+6y^{2}-y-30 by y^{2}+y-6.
  1. Divide the leading terms: y^{3}\div y^{2}=y. Multiply the divisor by y and subtract the product.
    \begin{aligned}y(y^{2}+y-6)&=y^{3}+y^{2}-6y\\(y^{3}+6y^{2}-y-30)-(y^{3}+y^{2}-6y)&=5y^{2}+5y-30\end{aligned}
  2. Divide the leading terms: 5y^{2}\div y^{2}=5. Multiply the divisor by 5 and subtract the product.
    \begin{aligned}5(y^{2}+y-6)&=5y^{2}+5y-30\\(5y^{2}+5y-30)-(5y^{2}+5y-30)&=0\end{aligned}
  3. The remainder has lower degree than the divisor (or is zero), so the division stops.
    \text{quotient}=y+5,\qquad\text{remainder}=0
  1. The remainder is zero and the quotient is y+5, so the third factor is y+5.
    y^{3}+6y^{2}-y-30=(y-2)(y+3)(y+5)
  2. Check with the last-term rule.
    (-2)(+3)(+5)=-30
y+5

Book answer. y+5

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 101 mark
In signal processing, P(x) represents a signal. If P(-1)=0, then the signal has:
  1. AA zero at x=-1Answer
  2. BUnstablity
  3. CPeriodicity
  4. DError
Why

If P(-1)=0, then x=-1 is a zero (root) of P(x), because a zero of a polynomial is a number a with P(a)=0. By the Factor Theorem, (x+1) is then a factor of P(x).

Option B, C and D describe the signal, not the polynomial: the value P(-1)=0 says nothing about stability, periodicity or error. The only statement it proves is that P has a zero at x=-1.

2026 Annual I, Set B, MCQ 141 mark
In polynomial division, degree of remainder is:
  1. AGreater
  2. BEqual
  3. CLess than divisorAnswer
  4. DInfinite
Why

In the division algorithm \text{dividend}=\text{divisor}\times\text{quotient}+\text{remainder}, the division stops only when the remainder is 0 or has a degree smaller than the degree of the divisor. So the remainder is always of lower degree than the divisor.

Options A and B would mean the remainder could still be divided by the divisor, so the division would not be finished. Option D is meaningless for a polynomial.

2026 Annual I, Set B, MCQ 151 mark
If (x-2) is factor of f(x), then f(2)=
  1. A-2
  2. B0Answer
  3. C1
  4. D2
Why

By the Factor Theorem, (x-a) is a factor of f(x) exactly when f(a)=0. Here a=2, so f(2)=0.

Option A, -2, is the tempting wrong choice: it would come from f(-2), which belongs to the factor (x+2). The factor (x-2) vanishes at x=2, not at x=-2.

2025 Annual I, Set B, Q.2 (iv) (a)
If P(x)=x^4-6x^3+11x^2-6x, divide P(x) by (x-1) using synthetic division.
Solution
  1. Write the coefficients of P(x)=x^{4}-6x^{3}+11x^{2}-6x in order of powers, putting 0 for the missing constant term. The divisor is x-1, so the synthetic divisor is c=1.
    1,\ -6,\ 11,\ -6,\ 0
  2. Bring down the first coefficient. Multiply by 1, write the product under the next coefficient and add. Repeat to the end.
Synthetic division by x-1 (first row: coefficients; second row: products; last row: sums).
11-611-60
Multiply by 11-560
Add1-5600

The last number is the remainder, 0, so x-1 is a factor (Factor Theorem). The other numbers are the coefficients of the quotient, which has degree one less than P.

x^{4}-6x^{3}+11x^{2}-6x=(x-1)(x^{3}-5x^{2}+6x)
\text{quotient }x^{3}-5x^{2}+6x,\ \text{remainder }0
2025 Annual I, Set B, Q.2 (iv) (b)
If P(x)=x^4-6x^3+11x^2-6x, solve the resulting depressed equation.
Solution
  1. The depressed equation is the quotient from part (a) put equal to zero.
    x^{3}-5x^{2}+6x=0
  2. Take out the common factor x.
    x(x^{2}-5x+6)=0
  3. Factorise the quadratic: two numbers with product 6 and sum -5 are -2 and -3.
    x(x-2)(x-3)=0
  4. Put each factor equal to zero.
    x=0,\quad x=2,\quad x=3

Together with the factor x-1 from part (a), all the zeros of P(x) are 0,1,2,3, and P(x)=x(x-1)(x-2)(x-3).

x=0,\ 2,\ 3
2026 Annual I, Set B, Q.2 (iv) OR4 marks
Find all roots of: x^3-3x^2-4x+12 by Factor theorem.
Solution

Let f(x)=x^{3}-3x^{2}-4x+12. Try the divisors of the last term 12 as values of x until f(x)=0.

  1. Try x=2.
    f(2)=8-12-8+12=0
  2. Since f(2)=0, by the Factor Theorem (x-2) is a factor. Divide by x-2 with synthetic division, c=2, coefficients 1,\ -3,\ -4,\ 12.
Synthetic division of x^{3}-3x^{2}-4x+12 by x-2.
21-3-412
Multiply by 22-2-12
Add1-1-60
  1. The remainder is 0 and the quotient is x^{2}-x-6.
    x^{3}-3x^{2}-4x+12=(x-2)(x^{2}-x-6)
  2. Factorise the quadratic: two numbers with product -6 and sum -1 are -3 and 2.
    x^{2}-x-6=(x-3)(x+2)
  3. Write the complete factorisation and put each factor equal to zero.
    f(x)=(x-2)(x-3)(x+2)=0\ \Rightarrow\ x=2,\ 3,\ -2

The roots are 2,\ 3 and -2.

x=2,\ 3,\ -2
2026 Annual I, Set B, Q.2 (ix)4 marks
Determine whether (x+2) is a factor of: f(x)=x^3+2x^2-4x-8.
Solution

By the Factor Theorem, (x+2)=(x-(-2)) is a factor of f(x) if and only if f(-2)=0.

  1. Substitute x=-2.
    f(-2)=(-2)^{3}+2(-2)^{2}-4(-2)-8
  2. Evaluate each term.
    f(-2)=-8+8+8-8
  3. Add.
    f(-2)=0

Since f(-2)=0, (x+2) is a factor of f(x).

f(-2)=0,\ \text{so } (x+2)\text{ is a factor}
2025 Annual I, Set B, Q.2 (xi) OR4 marks
If P(x)=ax^3+bx^2+2x-1 leaves remainder 5 when divided by (x-1) and leaves remainder 3 when divided by (x+1). Find values of a and b using remainder theorem.
Solution
  1. Remainder Theorem: the remainder on dividing by x-1 is P(1). It is 5.
    P(1)=a+b+2-1=5\;\Longrightarrow\;a+b=4
  1. The remainder on dividing by x+1=x-(-1) is P(-1). It is 3.
    P(-1)=-a+b-2-1=3\;\Longrightarrow\;-a+b=6
  1. Add the two equations to eliminate a.
    2b=10\;\Longrightarrow\;b=5
  2. Substitute in a+b=4.
    a=4-5=-1
  3. Check: P(x)=-x^{3}+5x^{2}+2x-1, so P(1)=-1+5+2-1=5 and P(-1)=1+5-2-1=3.
a=-1,\ b=5
2025 Annual I, Set B, Q.3 (d)
If P(x)=2x^3+x^2-13x+6, write complete factorized form of P(x) and confirm by expanding.
Solution
  1. From part (b), the complete factorised form is
    P(x)=(x+3)(2x-1)(x-2)
  2. Confirm by expanding. First multiply the last two brackets.
    (2x-1)(x-2)=2x^{2}-4x-x+2=2x^{2}-5x+2
  3. Multiply by (x+3).
    (x+3)(2x^{2}-5x+2)=2x^{3}-5x^{2}+2x+6x^{2}-15x+6
  4. Collect like terms.
    =2x^{3}+x^{2}-13x+6=P(x)

The expansion gives back P(x), so the factorisation is confirmed.

P(x)=(x+3)(2x-1)(x-2)
2025 Annual I, Set B, Q.3 (a)
If P(x)=2x^3+x^2-13x+6, verify that (x+3) is a factor of P(x).
Solution
  1. By the Factor Theorem, (x+3)=(x-(-3)) is a factor of P(x) if P(-3)=0.
  2. Substitute x=-3.
    P(-3)=2(-27)+9-13(-3)+6=-54+9+39+6
  3. Add.
    P(-3)=0

Since P(-3)=0, (x+3) is a factor of P(x). Hence verified.

P(-3)=0
2025 Annual I, Set B, Q.3 (b)
If P(x)=2x^3+x^2-13x+6, factorize P(x) using synthetic division.
Solution

Since (x+3) is a factor, divide by x+3 with synthetic divisor c=-3. Coefficients: 2,\ 1,\ -13,\ 6.

Synthetic division of 2x^{3}+x^{2}-13x+6 by x+3.
-321-136
Multiply by -3-615-6
Add2-520
  1. The remainder is 0 and the quotient is 2x^{2}-5x+2.
    2x^{3}+x^{2}-13x+6=(x+3)(2x^{2}-5x+2)
  2. Factorise the quadratic: product 2\times2=4, sum -5, so split -5x=-4x-x.
    2x^{2}-4x-x+2=2x(x-2)-1(x-2)=(x-2)(2x-1)
  3. Collect the factors.
    P(x)=(x+3)(2x-1)(x-2)
P(x)=(x+3)(2x-1)(x-2)
2025 Annual I, Set B, Q.3 (c)
If P(x)=2x^3+x^2-13x+6, find all the roots of P(x).
Solution

The roots of P(x)=0 are the zeros of P(x). Use the factorisation of part (b).

  1. Put the factorised form equal to zero.
    (x+3)(2x-1)(x-2)=0
  2. Put each factor equal to zero.
    x+3=0,\quad 2x-1=0,\quad x-2=0
  3. Solve.
    x=-3,\quad x=\dfrac12,\quad x=2
x=-3,\ \dfrac12,\ 2
2026 Annual I, Set B, Q.6 (a)3+5 printed for parts (a), (b) of the 8-mark Q.6 = 3 marks
For what value of k the remainder of 2x^3-kx^2+16x+12 will be zero if the polynomial is divided by x-2.
Solution

By the Remainder Theorem, the remainder on division by x-2 is P(2). The remainder is zero, so P(2)=0.

  1. Let P(x)=2x^{3}-kx^{2}+16x+12 and substitute x=2.
    P(2)=2(8)-k(4)+16(2)+12=16-4k+32+12
  2. Simplify.
    P(2)=60-4k
  3. Put the remainder equal to zero and solve for k.
    60-4k=0\ \Rightarrow\ k=15
k=15
2026 Annual I, Set B, Q.6 (b)3+5 printed for parts (a), (b) of the 8-mark Q.6 = 5 marks
Factorize the polynomial 2x^3-kx^2+16x+12 completely (using the value of k found in part (a)).
Solution

From part (a), k=15, so P(x)=2x^{3}-15x^{2}+16x+12. Since P(2)=0, (x-2) is a factor. Divide by x-2 with synthetic division, c=2, coefficients 2,\ -15,\ 16,\ 12.

Synthetic division of 2x^{3}-15x^{2}+16x+12 by x-2.
22-151612
Multiply by 24-22-12
Add2-11-60
  1. The remainder is 0 and the quotient is 2x^{2}-11x-6.
    2x^{3}-15x^{2}+16x+12=(x-2)(2x^{2}-11x-6)
  2. Factorise the quadratic: product 2\times(-6)=-12, sum -11, so split -11x=-12x+x.
    2x^{2}-12x+x-6=2x(x-6)+1(x-6)=(x-6)(2x+1)
  3. Collect the factors.
    P(x)=(x-2)(x-6)(2x+1)
P(x)=(x-2)(x-6)(2x+1)
2026 Annual I, Set D, MCQ 91 mark
If a cubic polynomial is divided by a linear polynomial (containing same variable), the degree of remainder will be:
  1. A0Answer
  2. B1
  3. C2
  4. D3
Why

The remainder has a degree smaller than the divisor. The divisor is linear (degree 1), so the remainder has degree less than 1, that is degree 0: it is a constant (possibly 0). For example P(x) divided by x-c leaves the constant P(c).

Option B, degree 1, is the tempting wrong choice because it is the degree of the divisor; a remainder of degree 1 could still be divided by the linear divisor.

2026 Annual I, Set D, MCQ 101 mark
Using factor theorem if f(a)=0 then:
  1. Af(a) is a factor
  2. Ba is factor
  3. C(x-a) is factorAnswer
  4. D(x+a) is factor
Why

The Factor Theorem says: if f(a)=0, then (x-a) is a factor of f(x) (and conversely).

Option D, (x+a), is the usual trap: it is the factor when f(-a)=0. Options A and B are not factors at all: f(a) is a number and a is a number, while a factor of a polynomial is itself a polynomial.

2025 Annual I, Set D, MCQ 101 mark
The division of 2x^3-5x^2+3x-1 by (x-2) gives remainder:
  1. A-3
  2. B0
  3. C5
  4. D1Answer
Why

By the Remainder Theorem the remainder on dividing by x-2 is P(2).

P(2)=2(8)-5(4)+3(2)-1=16-20+6-1=1

The remainder is 1, not 0, so x-2 is not a factor (option B is wrong). Substituting x=-2 instead of x=2 would be the mistake of dividing by x+2.

2026 Annual I, Set D, Q.2 (iv)4 marks
Find the value(s) of K if x-1 is a factor of: f(x)=x^3+K^2-2K+1
Solution

By the Factor Theorem, (x-1) is a factor of f(x) if and only if f(1)=0.

  1. Substitute x=1 in f(x)=x^{3}+K^{2}-2K+1.
    f(1)=1+K^{2}-2K+1=K^{2}-2K+2
  2. Put f(1)=0 to get a quadratic equation in K.
    K^{2}-2K+2=0
  3. Use the quadratic formula with a=1,\ b=-2,\ c=2.
    K=\dfrac{2\pm\sqrt{4-8}}{2}=\dfrac{2\pm2i}{2}
  4. Simplify.
    K=1\pm i

The discriminant b^{2}-4ac=-4 is negative, so there is no real value of K; the values are the complex numbers 1+i and 1-i.

K=1+i\ \text{ or }\ K=1-i

Note. Solved exactly as printed. The printed f(x) has no x in the terms with K, so f(1)=0 gives complex K. If the paper meant x^{3}+Kx^{2}-2K+1, then f(1)=2-K=0 gives K=2.

2025 Annual I, Set D, Q.2 (iv) (a)
If P(x)=3x^3-5x^2-16x+12, use factor theorem to verify that (x+2) is a factor of P(x).
Solution
  1. By the Factor Theorem, (x+2)=(x-(-2)) is a factor if P(-2)=0.
  2. Substitute x=-2 in P(x)=3x^{3}-5x^{2}-16x+12.
    P(-2)=3(-8)-5(4)-16(-2)+12=-24-20+32+12
  3. Add.
    P(-2)=0

So (x+2) is a factor of P(x). Hence verified.

P(-2)=0
2025 Annual I, Set D, Q.2 (iv) (b)
If P(x)=3x^3-5x^2-16x+12, apply synthetic division to factorize P(x) completely.
Solution

Divide by the factor x+2 with synthetic divisor c=-2. Coefficients: 3,\ -5,\ -16,\ 12.

Synthetic division of 3x^{3}-5x^{2}-16x+12 by x+2.
-23-5-1612
Multiply by -2-622-12
Add3-1160
  1. Remainder 0, quotient 3x^{2}-11x+6.
    P(x)=(x+2)(3x^{2}-11x+6)
  2. Factorise the quadratic: product 3\times6=18, sum -11, so split -11x=-9x-2x.
    3x^{2}-9x-2x+6=3x(x-3)-2(x-3)=(x-3)(3x-2)
  3. Collect the factors.
    P(x)=(x+2)(x-3)(3x-2)
P(x)=(x+2)(x-3)(3x-2)
2026 Annual I, Set D, Q.2 (xi) OR4 marks
Factorize the polynomial: x^3-4x^2+x+6.
Solution

Let f(x)=x^{3}-4x^{2}+x+6. Try divisors of the last term 6.

  1. Try x=-1.
    f(-1)=-1-4-1+6=0
  2. Since f(-1)=0, (x+1) is a factor. Divide by x+1 with synthetic division, c=-1, coefficients 1,\ -4,\ 1,\ 6.
Synthetic division of x^{3}-4x^{2}+x+6 by x+1.
-11-416
Multiply by -1-15-6
Add1-560
  1. The remainder is 0 and the quotient is x^{2}-5x+6.
    x^{3}-4x^{2}+x+6=(x+1)(x^{2}-5x+6)
  2. Factorise the quadratic: product 6, sum -5, so the numbers are -2 and -3.
    x^{2}-5x+6=(x-2)(x-3)
  3. Collect the factors.
    f(x)=(x+1)(x-2)(x-3)
f(x)=(x+1)(x-2)(x-3)
2025 Annual I, Set D, Q.2 (xi) OR (a)
A checksum polynomial C(x)=x^4+px^2+qx+6 is used to detect errors in transmitted data. If C(1)=10 and C(-1)=4, set equations for p and q.
Solution
  1. Put x=1 in C(x)=x^{4}+px^{2}+qx+6 and use C(1)=10.
    1+p+q+6=10\;\Longrightarrow\;p+q=3
  2. Put x=-1 and use C(-1)=4.
    1+p-q+6=4\;\Longrightarrow\;p-q=-3
p+q=3,\quad p-q=-3
2025 Annual I, Set D, Q.2 (xi) OR (b)
For the checksum polynomial C(x)=x^4+px^2+qx+6, solve the equations to find values of p and q.
Solution
  1. Add the equations p+q=3 and p-q=-3.
    2p=0\;\Longrightarrow\;p=0
  2. Substitute in p+q=3.
    q=3
  3. Check: C(x)=x^{4}+3x+6, so C(1)=10 and C(-1)=1-3+6=4.
p=0,\ q=3
2026 Annual I, Set D, Q.3 (a)
A company's profit over the last few years can be modeled by the polynomial P(x)=x^3+6x^2-x-6. If x represents years since 2010, calculate the company's profit in 2012 by dividing P(x) by x-2 and finding the remainder.
Solution

The year 2012 is x=2 years after 2010. By the Remainder Theorem, the remainder on dividing P(x) by x-2 is P(2), which is the profit in 2012. Divide by x-2 with synthetic division, c=2, coefficients 1,\ 6,\ -1,\ -6.

Synthetic division of x^{3}+6x^{2}-x-6 by x-2.
216-1-6
Multiply by 221630
Add181524
  1. Read the quotient and the remainder from the last row.
    P(x)=(x-2)(x^{2}+8x+15)+24
  2. The remainder is the value of P at x=2, which is the profit in 2012.
    P(2)=24

The company's profit in 2012 is 24 (in the units used by the model P(x)).

\text{Remainder}=P(2)=24
2025 Annual I, Set D, Q.3 (a)
A polynomial P(x)=x^4+ax^3-5x^2+bx+c leaves remainder 0 when divided by (x-0). Use remainder theorem to find values of a, b and c.
Solution
  1. Dividing by x-0=x leaves remainder P(0) (Remainder Theorem). It is 0.
    P(0)=c=0

So c=0. The values of a and b come from parts (b) and (c) together; this part gives c=0.

c=0
2026 Annual I, Set D, Q.3 (b)
Verify your answer (the remainder of P(x)=x^3+6x^2-x-6 divided by x-2) by long division.
Solution

Divide x^{3}+6x^{2}-x-6 by x-2 by long division.

  1. Divide the leading terms: x^{3}\div x=x^{2}. Multiply the divisor by x^{2} and subtract.
    \begin{aligned}x^{2}(x-2)&=x^{3}-2x^{2}\\(x^{3}+6x^{2}-x-6)-(x^{3}-2x^{2})&=8x^{2}-x-6\end{aligned}
  2. Divide the leading terms: 8x^{2}\div x=8x. Multiply the divisor by 8x and subtract.
    \begin{aligned}8x(x-2)&=8x^{2}-16x\\(8x^{2}-x-6)-(8x^{2}-16x)&=15x-6\end{aligned}
  3. Divide the leading terms: 15x\div x=15. Multiply the divisor by 15 and subtract.
    \begin{aligned}15(x-2)&=15x-30\\(15x-6)-(15x-30)&=24\end{aligned}
  4. The remainder 24 has lower degree than the divisor, so the division stops.
    \text{quotient}=x^{2}+8x+15,\qquad\text{remainder}=24
  1. Check with dividend = divisor \times quotient + remainder.
    (x-2)(x^{2}+8x+15)+24=x^{3}+8x^{2}+15x-2x^{2}-16x-30+24=x^{3}+6x^{2}-x-6

The remainder by long division is 24, the same as P(2)=24 in part (a). Verified.

\text{Remainder}=24=P(2)
2025 Annual I, Set D, Q.3 (b)
A polynomial P(x)=x^4+ax^3-5x^2+bx+c leaves remainder -3 when divided by (x-2). Use remainder theorem to find values of a, b and c.
Solution
  1. The remainder on dividing by x-2 is P(2). It is -3.
    P(2)=16+8a-20+2b+c=-3
  2. Simplify.
    8a+2b+c=1

Using c=0 from part (a): 8a+2b=1.

8a+2b+c=1
2025 Annual I, Set D, Q.3 (c)
A polynomial P(x)=x^4+ax^3-5x^2+bx+c leaves remainder -4 when divided by (x+1). Use remainder theorem to find values of a, b and c.
Solution
  1. The remainder on dividing by x+1=x-(-1) is P(-1). It is -4.
    P(-1)=1-a-5-b+c=-4
  2. Simplify.
    -a-b+c=0
  3. With c=0 from (a): b=-a. Put this in 8a+2b=1 from (b).
    8a-2a=1\;\Longrightarrow\;a=\dfrac16
  4. Then b=-a.
    b=-\dfrac16

Check: P(2)=16+\dfrac86-20-\dfrac26+0=-4+1=-3, and P(-1)=1-\dfrac16-5+\dfrac16=-4.

a=\dfrac16,\ b=-\dfrac16,\ c=0

Exam focus

  1. Factorise a cubic completely: find one zero by trial (divisors of the last term), divide with synthetic division, then factorise the quadratic. (problem)
    Asked in both years: 2026 Set B Q.2 (iv) OR (roots of x^{3}-3x^{2}-4x+12), Set B Q.6 (b) (5 marks, 2x^{3}-15x^{2}+16x+12), Set D Q.2 (xi) OR (x^{3}-4x^{2}+x+6); 2025 Set B Q.3 (b) and Set D Q.2 (iv) (b). Book: Exercise 5.2, and Q6 of it is the 2026 Set B Q.6 (b) question.
    Go to answer
  2. Find an unknown coefficient (k, K, p, q) from a given zero, factor or remainder using P(c)=r. (problem)
    2026 Set B Q.6 (a) (k=15, 3 marks) and Set D Q.2 (iv) (x-1 a factor of the given f(x)); 2025 Set B Q.2 (xi) OR and Set D Q.3 (8 marks, three remainder conditions). Book: Exercise 5.1 Q5 and Q6.
    Go to answer
  3. Verify that x-c is a factor by showing f(c)=0. (problem)
    2026 Set B Q.2 (ix) (x+2 a factor of x^{3}+2x^{2}-4x-8), 2025 Set B Q.3 (a) and Set D Q.2 (iv) (a). A 4-mark question done by one substitution; remember x+2 means c=-2.
    Go to answer
  4. Remainder from a word problem: divide the model polynomial by x-c and read the remainder as P(c); verify by long division. (problem)
    2026 Set D Q.3 (a), (b) (8 marks, profit model x^{3}+6x^{2}-x-6 divided by x-2, remainder 24). Needs both synthetic division and long division with the check dividend = divisor \times quotient + remainder. Book: Section 5.6 and Exercise 5.3.
    Go to answer
  5. Short theory MCQs: the remainder of a division has a degree less than the divisor; f(a)=0 means (x-a) is a factor. (mcq)
    2026 Set B MCQ 14 and MCQ 15, Set D MCQ 9 and MCQ 10; 2025 Set B MCQ 10 and Set D MCQ 10. Six MCQs in two years, so learn the Remainder and Factor Theorem statements exactly. Book: Review Exercise 5 MCQs (iv), (v), (vi).
    Go to answer
  6. Synthetic division by x-c and solving the depressed (quotient) equation. (problem)
    2025 Set B Q.2 (iv) (a), (b): x^{4}-6x^{3}+11x^{2}-6x divided by x-1. The same table is the working step in every 2026 factorisation; write 0 for any missing term.
    Go to answer
  7. Polynomial long division, writing dividend = quotient \times divisor + remainder. (problem)
    2026 Set D Q.3 (b) asked for it directly. Review Exercise 5 Q2, Q3 and Q7 practise it, and it underlies the synthetic method.
    Go to answer
  8. Word problems on volumes and areas that give a polynomial and one side, with the other sides found by factorising. (problem)
    Not in the 2025 or 2026 papers; Exercise 5.3 (Q1 to Q6) and Exercise 5.1 Q10 practise it, so keep it as a practice set.
    Go to answer

Mark this unit as done to fill the Class 11 Mathematics ring.

Next, unit 6
Permutation and combination

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