Create New Account
Create New Account

Sign in to Innings2

or

New Account     Reset Password     

Polynomials

    Exercise 2.3

Powered by Innings 2

Reset Progress

This will delete your progress and chat data for all chapters in this course, and cannot be undone!

Glossary

Select one of the keywords on the left…

9th class > 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 =