8 - Factor Theorem

In this post, you can check the solutions to the exercises of Factor Theorem of Algebra Unit of Readmore Optional Mathematics by DR Simkhada. It consists of complete answers to all odd questions mentioned in the text book.

Here are a few things that you need to consider before checking the solutions:
  1. Factor theorem states, "If f(x) be a polynomial of degree n where n>0 is divided by a binomial g(x) = (x -a) such that a is a real number, then for f(a) = 0, g(x) is a factor of f(x).
  2. When we have (px - q), a = q/p.
Now, you can check the solutions below. But, make sure to check the description after the solution to know more about this textbook series, author, and the publisher.


1. In each of the following, use the factor theorem to check whether or not the polynomial g(x) is the factor of the polynomial f(x).

a) f(x) = x⁴ + x³ - x² - x - 18; g(x) = x -2
Solution:
Comparing (x -2) with (x -a), we get, a = 2
Finding the remainder using remainder theorem,
f(a) = f(2) = 2⁴ + 2³ - 2² - 2 - 18
= 16 + 8 - 4 - 2 - 18
= 0
f(a) = 0. Thus, according to factor theorem, g(x) is a factor of f(x).

c) f(x) = x³ + x² + x + 1; g(x) = x - 1
Solution:
Comparing (x -1) with (x - a), we get, a = 1
Finding the remainder using remainder theorem,
f(a) = f(1) = 1³ + 1² + 1 + 1
= 4
f(a) ≠ 0. Thus, according to factor theorem, g(x) is not a factor of f(x).

2a) For what value of k, the polynomial x³ + 4x² - kx + 8 is exactly divisible by (x -2)?
Solution:
Let f(x) = x³ + 4x² - kx + 8.
According to question, (x -2) is a factor of f(x). Comparing (x -2) with (x -a), we get, a = 2
Using factor theorem,
f(a) = 0
or, f(2) = 0
or, 2³ + 4(2)² - k(2) + 8 = 0
or, 8 + 16 - 2k + 8 = 0
or, 32 = 2k
or, k = 32/2
So, k = 16

2c) For what value of b, the polynomial x³ - 3x² + bx - 6 is exactly divisible by (x -3)?
Solution:
Let f(x) = x³ - 3x² + bx - 6
According to question, (x -3) is a factor of f(x). Comparing (x -3) with (x -a), we get, a = 3
Using factor theorem,
f(a) = 0
or, f(3) = 0
or, 3³ - 3(3)² + b(3) - 6 = 0
or, 27 - 27 + 3b - 6 = 0
or, 3b = 6
or, b = 6/3
So, b = 2

3a) If x-2 is a factor of x⁵ - 3x⁴ - kx³ + 3kx² +2kx +4, find the value of k.
Solution:
Let f(x) = x⁵ - 3x⁴ - kx³ + 3kx² + 2kx +4
Comparing (x -2) with (x -a), we get, a = 2
According to factor theorem,
f(a) = 0
or, f(2) = 0
or, 2⁵ - 3(2)⁴ - k(2)³ + 3k(2)² + 2k(2) + 4 = 0
or, 32 - 48 - 8k + 12k + 4k + 4 = 0
or, 8k - 12 = 0
or, 8k = 12
or, k = 12/8
So, k = 3/2

3c) If x - k is a factor of x² + 6x - kx - 24, find the value of k.
Solution:
Let f(x) = x² + 6x - kx - 24
Comparing (x - k) with (x -a), we get, a = k
According to factor theorem,
f(a) = 0
or, f(k) = 0
or, (k)² + 6(k) - k(k) - 24 = 0
or, k² + 6k - k² - 24 = 0
or, 6k - 24 = 0
or, 6k = 24
or, k = 24/6
So, k = 4

3e) If x+3 is a factor of x³ - (k-1)x² + kx + 54, find the value of k.
Solution:
Let f(x) = x³ - (k-1)x² + kx + 54
Comparing (x +3) with (x -a), we get, a = -3
According to factor theorem,
f(a) = 0
or, f(-3) = 0
or, (-3)³ - (k-1)(-3)² + k(-3) + 54 = 0
or, -27 - (k-1)9 -3k + 54 = 0
or, - (k -1)9 - 3k + 27 = 0
or, 27 = (k -1)9 + 3k
or, 27 = 9k - 9 + 3k
or, 27 + 9 = 12k
or, 36 = 12k
or, k = 36/12
So, k = 3

4. Factorise:

a) x² + 8x + 15
Solution:
x² + 8x + 15
= x² + (3+5)x + 15
= x² + 3x + 5x + 15
= x(x +3) + 5(x +3)
= (x + 3)(x + 5)

c) x² + 2x - 15
Solution:
x² + 2x - 15
= x² + (5 - 3)x - 15
= x² + 5x - 3x - 15
= x(x + 5) - 3(x + 5)
= (x - 3)(x + 5)

e) 3x² + 5x + 2
Solution:
3x² + 5x + 2
= 3x² + (3+2)x + 2
= 3x² + 3x + 2x + 2
= 3x(x +1) + 2(x + 1)
= (3x +2)(x + 1)
= (x + 1)(3x + 2)

g) x³ - 2x² - 5x + 6
Solution:
Let f(x) = x³ - 2x² - 5x + 6
According to factor theorem, if f(a) = 0, (x -a) is a factor.
Testing a = 1
f(1) = 1³ - 2(1)² - 5(1) + 6
= 1 - 2 - 5 + 6
= 7 - 7
= 0
So, (x - 1) is a factor.
Now,
Using synthetic division,
1 | 1 -2 -5  6
   |      1 -1 -6
   ___________
     1 -1 -6   ×
We get,
x³ - 2x² - 5x + 6
= (x -1)(x² - x -6)
= (x -1)(x² - (3-2)x -6)
= (x -1)(x² - 3x +2x -6)
= (x -1){x(x -3) + 2(x -3)}
= (x -1)(x +2)(x -3)

i) x³ - x² - 14x + 24
Solution:
Let f(x) = x³ - 2x² - 5x + 6
According to factor theorem, if f(a) = 0, (x -a) is a factor.
Testing a = 2
f(1) = 2³ - (2)² - 14(2) + 24
= 8 - 4 - 28 + 24
= 32 - 32
= 0
So, (x - 2) is a factor.
Now,
Using synthetic division,
2 | 1 -1 -14  24
   |      2    2 -24
   ______________
     1  1  -12  ×
We get,
x³ - x² - 14x + 24
= (x -2)(x² + x -12)
= (x -2){x² + (4-3)x - 12}
= (x -2){x² + 4x - 3x -12}
= (x -2){x(x +4) - 3(x +4)}
= (x -2)(x -3)(x +4)

5a) Find the values of 'a' and 'b'. If (x -1) and (x -2) are factors of x³ + ax² + bx - 6.
Solution:
Let f(x) = x³ + ax² + bx - 6
According to factor theorem,
If (x -a) is a factor then f(a) = 0
In condition I,
Comparing (x -1) with (x -a), we get, a = 1
f(1) = 0
or, (1)³ + a(1)² + b(1) - 6 = 0
or, 1 + a + b - 6 = 0
or, a + b - 5 = 0
or, a + b = 5
or, b = 5 - a  ---(i)
In condition II,
Comparing (x -2) with (x -a), we get, a = 2
f(2) = 0
or, (2)³ + a(2)² + b(2) - 6 = 0
or, 8 + 4a + 2b - 6 = 0
or, 4a + 2b + 2 = 0
or, 2(2a + b + 1) = 0
or, 2a + b + 1 = 0
or, 2a + b = - 1
or, b = -1 - 2a ---(ii)
From equations (i) and (ii), we get,
5 - a = - 1 - 2a
or, 5 + 1 = a - 2a
or, 6 = - a
So, a = -6
Put value of a in equation (i), we get,
b = 5 - (-6) = 5 + 6 = 11
So, (a,b) = (-6,11)

5c) If x³ + ax² + bx + 6 has x-2 as a factor and leaves a remainder 3 when divided by (x -3), find a and b.
Solution:
Let f(x) = x³ + ax² + bx + 6
In condition I,
Comparing (x -2) with (x -a), we get, a = 2
According to factor theorem,
f(a) = f(2) = 0
or, (2)³ + a(2)² + b(2) + 6 = 0
or, 8 + 4a + 2b + 6 = 0
or, 4a + 2b + 14 = 0
or, 2(2a + b + 7) = 0
or, 2a + b + 7 = 0
or, b = - (7 + 2a) ---(i)
In condition II,
Comparing (x -3) with (x -a), we get, a = 3
According to remainder theorem and question,
f(a) = f(3) = 3
or, (3)³ + a(3)² + b(3) + 6 = 3
or, 27 + 9a + 3b + 6 = 3
or, 9a + 3b + 33 = 3
or, 9a + 3b + 30 = 0
or, 3(3a + b + 10) = 0
or, 3a + b + 10 = 0
or, b = - (10 + 3a) ---(ii)
From equations (i) and (ii), we get,
- (7 + 2a) = - (10 + 3a)
or, 7 + 2a = 10 + 3a
or, 2a - 3a = 10 - 7
or, -a = 3
So, a = -3
Put value of a in equation (ii), we get,
b = - {10 + 3(-3)} = -{10 - 9} = -1
So, (a,b) = (-3,-1)

6a) What must be added to $x^3 - 6x^2 +11x -8$ to make a polynomial having a factor (x -3)$?

Solution:

For the above condition, we need to find the remainder using remainder theorem and then add the additive inverse of the remainder to the given function.

Let p(x) = x³ - 6x² + 11x -8

Comparing (x - 3) with (x - a), we get,
a = 3

Using remainder theorem,
p(a) = $3³ - 6(3)² + 11(3) - 8$

$= 27 - 54 + 33 - 8$

$= -2$

Reminder is - 2

Additive inverse of -2 is +2.

Hence, 2 should be added to p(x) to make it perfectly divisible by (x -3).


About Readmore Optional Mathematics Book 10

Author: D. R. Simkhada
Editor: I. R. Simkhada

Readmore Publishers and Distributors
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Factor Theorem - Solved Exercises | Class 10 Readmore Optional Mathematics is a collection of the solutions related to exercises of Factor Theorem from the Algebra chapter for Nepal's Secondary Education Examination (SEE) appearing students.