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Chapter 2: Polynomials

    Introduction

    Polynomials in One Variable

    Zeroes of a Polynomial

    Factorisation of Polynomials

    Algebraic Identities

    Exercise 2.1

    Exercise 2.2

    Exercise 2.3

    Exercise 2.4

    Summary

    Enhanced Curriculum Support

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Chapter 2: Polynomials > Exercise 2.3

Exercise 2.3

  1. Determine which of the following polynomials has (x + 1) a factor.

Instructions

(i) x3+x2+x+1
x + 1 = 0 which gives us x = .
Substituting −1 in equation.
x3+x2+x+1 = −13+−12+−1+1 = .
We have found the answer.
(ii)x4+x3+x2+x+1
x+1=0 gives us x = .
Substituting −1 in equation.
x4+x3+x2+x+1 = −14+−13+−12+−1+1 =
We have found the answer.
(iii)x4+3x3+3x2+x+1
x+1=0 gives us x = .
Substituting −1 in equation.
x4+3x3+3x2+x+1 = −14+3−13+3−12+−1+1 =
We have found the answer.
(iv) x3−x2−2+2x+2
x+1=0 gives us x = .
Substituting −1 in equation.
x3−x2−2+2x+2 = −13−−12−2+2−1+2 =
We have found the answer.
  1. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases.

Instructions

(i)2x3+x2−2x−1, gx=x+1
g(x) = 0 gives us x+1 = 0 i.e. x =
Substituting −1 in the equation.
p−1=2−13+−12−2−1−1 = .
Thus, g(x) is a factor of p(x).
(ii)x3+3x2+3x+1, gx=x+2
g(x) = 0 gives us x+2 = 0 i.e. x =
Substituting −2 in the equation.
p−2 = −23+3−22+3−2+1 = .
Thus, g(x) is not a factor of p(x).
(iii)x3−4x2+x+6, gx=x−3
g(x) = 0 gives us x−3 = 0 i.e. x =
Substituting 3 in the equation.
p(3) = 33−432+3+6 =
Thus, g(x) is a factor of p(x).
  1. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases.

Instructions

If x−1 is a factor of p(x), it means that
(i) p(x) = x2+x+k
p(1) = 12+1+k = 0 ⇒ k =
(ii) p(x) = 2x2+kx+2
p(1) = 212 + k(1) + 2 ⇒ k =
(iii) p(x) = kx2−2x+1
p(1) = k12−21+1 = 0 ⇒ k =
(iv) p(x) = kx2−3x+k
p(1) = k12−31+k = 0 ⇒ k =

4.Factorise:

Instructions

(i) 12x2−7x+1
12x2−7x+1 = 12x2 + 1
= =
(ii)2x2+7x+3
= 2x2+ +x+3 =
=
(iii)6x2+5x−6
= 6x2+9x - −6 =
=
(iv)3x2−x−4
= 3x2−4x+3x−4 =
=

5. Factorise:

Instructions

(i) x3-2x2-x + 2
Using trial and error method: substitute 0, ±1, ±2 etc. to find the first factor. We find that x = satisfies the equation.
This means that is a factor of x3−2x2−x+2.
Upon dividing x3−2x2−x+2 by x+1, we get the quotient:
Further factorising x2−3x+2, we get:
Thus, x3−2x2−x+2 =
(ii)x3-3x2-9x-5
The factors of 5 are ±1 and ±5. Substituting in the equation, we get that x = satisfies the equation.
Thus, is a factor of x3−3x2−9x−5.
Upon dividing x3−3x2−9x−5 by x−5 , we get the quotient:
Further factorising x2+2x+1 , we get:
Thus, x3−3x2−9x−5 =
(iii)x3+13x2+32x+20
The factors of 20 are ±1,±2,±4,±5 and ±10. Substituting in the equation, we get that x = satisfies the equation.
Thus, is a factor of x3+13x2+32x+20.
Upon dividing x3+13x2+32x+20 by x+1 , we get the quotient:
Further factorising x2+12x+20 , we get:
Thus, x3+13x2+32x+20 =
(iv)2y3+y2-2y-1
Using trial and error method, substitute ±1,±2 and 0. Substituting in the equation, we get that y = satisfies the equation.
Thus, is a factor of 2y3+y2−2y−1.
Upon dividing 2y3+y2−2y−1 by y−1 , we get the quotient:
Further factorising 2y2+3y+1 , we get:
Thus, 2y3+y2−2y−1 =