ICSE Solutions for Selina Concise Chapter 8 Remainder and Factor Theorem Class 10 Maths
Exercise 8(A)
1. Find, in each case, the remainder when:
(i) x4 – 3x2 + 2x + 1 is divided by x – 1.
(ii) x3 + 3x2 – 12x + 4 is divided by x – 2.
(ii) x4 + 1 is divided by x + 1.
Solution
From remainder theorem, we know that when a polynomial f (x) is divided by (x – a), then the remainder is f(a).
(i) Given, f(x) = x4 – 3x2 + 2x + 1 is divided by x – 1
So, remainder = f(1) = (1)4 – 3(1)2 + 2(1) + 1 = 1 – 3 + 2 + 1 = 1
(ii) Given, f(x) = x3 + 3x2 – 12x + 4 is divided by x – 2
So, remainder = f(2) = (2)3 + 3(2)2 – 12(2) + 4 = 8 + 12 – 24 + 4 = 0
(iii) Given, f(x) = x4 + 1 is divided by x + 1
So, remainder = f(-1) = (-1)4 + 1 = 2
2. Show that:
(i) x – 2 is a factor of 5x2 + 15x – 50
(ii) 3x + 2 is a factor of 3x2 – x – 2
Solution
(x – a) is a factor of a polynomial f(x) if the remainder, when f(x) is divided by (x – a), is 0, i.e., if f(a) = 0.
(i) f(x) = 5x2 + 15x – 50
f(2) = 5(2)2 + 15(2) – 50
= 20 + 30 – 50
= 0
As the remainder is zero for x = 2
Thus, we can conclude that (x – 2) is a factor of 5x2 + 15x – 50
(ii) f(x) = 3x2 – x – 2
f(-2/3) = 3(-2/3)2 – (-2/3) – 2
= 4/3 + 2/3 – 2
= 2 – 2
= 0
As the remainder is zero for x = -2/3
Thus, we can conclude that (3x + 2) is a factor of 3x2 – x – 2
3. Use the Remainder Theorem to find which of the following is a factor of 2x3 + 3x2 – 5x – 6.
(i) x + 1
(ii) 2x – 1
(iii) x + 2
Solution
From remainder theorem we know that when a polynomial f (x) is divided by x – a, then the remainder is f(a).
Here, f(x) = 2x3 + 3x2 – 5x – 6
(i) f (-1) = 2(-1)3 + 3(-1)2 – 5(-1) – 6
= -2 + 3 + 5 – 6
= 0
⇒ Remainder is zero for x = -1
∴ (x + 1) is a factor of the polynomial f(x).
(ii) f(1/2) = 2(1/2)3 + 3(1/2)2 – 5(1/2) – 6
= ¼ + ¾ – 5/2 – 6
= -5/2 – 5
= -15/2
⇒ Remainder is not equals to zero for x = 1/2
∴ (2x – 1) is not a factor of the polynomial f(x).
(iii) f (-2) = 2(-2)3 + 3(-2)2 – 5(-2) – 6
= -16 + 12 + 10 – 6
= 0
⇒ Remainder is zero for x = -2
∴ (x + 2) is a factor of the polynomial f(x).
4. (i) If 2x + 1 is a factor of 2x2 + ax – 3, find the value of a.
(ii) Find the value of k, if 3x – 4 is a factor of expression 3x2 + 2x – k.
Solution
(i) Given, 2x + 1 is a factor of f(x) = 2x2 + ax – 3.
So, f(-1/2) = 0
⇒ 2(-1/2)2 + a(-1/2) – 3 = 0
⇒ ½ – a/2 – 3 = 0
⇒ 1 – a – 6 = 0
⇒ a = -5
(ii) Given, 3x – 4 is a factor of g(x) = 3x2 + 2x – k.
So, f(4/3) = 0
⇒ 3(4/3)2 + 2(4/3) – k = 0
⇒ 16/3 + 8/3 – k = 0
⇒ 24/3 = k
⇒ k = 8
5. Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression x3 + ax2 + bx – 12.
Solution
Here, f(x) = x3 + ax2 + bx – 12
Given, x – 2 and x + 3 both are the factors of f(x)
So,
f(2) and f(-3) both should be equal to zero.
f(2) = (2)3 + a(2)2 + b(2) – 12
⇒ 0 = 8 + 4a + 2b – 12
⇒ 0 = 4a + 2b – 4
⇒ 2a + b = 2 …. (1)
Now,
f(-3) = (-3)3 + a(-3)2 + b(-3) – 12
⇒ 0 = -27 + 9a – 3b – 12
⇒ 9a – 3b – 39 = 0
⇒ 3a – b = 13 …. (2)
Adding (1) and (2), we get,
5a = 15
Thus, a = 3
Putting the value of a in (1), we have
6 + b = 2
Thus, b = -4
6. Find the value of k, if 2x + 1 is a factor of (3k + 2)x3 + (k – 1).
Solution
Let take f(x) = (3k + 2)x3 + (k – 1)
Now, 2x + 1 = 0
x = -1/2
As, 2x + 1 is a factor of f(x) then the remainder should be 0.
f(-1/2) = (3k + 2)(-1/2)3 + (k – 1) = 0
⇒ 5k = 10 = 0
⇒ k = 2
7. Find the value of a, if x – 2 is a factor of 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8.
Solution
Given, f(x) = 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8 and x – 2 is a factor of f(x).
So, x – 2 = 0; x = 2
Hence, f(2) = 0
2(2)5 – 6(2)4 – 2a(2)3 + 6a(2)2 + 4a(2) + 8 = 0
⇒ 64 – 96 – 16a + 24a + 8a + 8 = 0
⇒ -24 + 16a = 0
⇒ 16a = 24
Thus, a = 3/2 = 1.5
8. Find the values of m and n so that x – 1 and x + 2 both are factors of x3 + (3m + 1) x2 + nx – 18.
Solution
Let f(x) = x3 + (3m + 1) x2 + nx – 18
Given, (x – 1) and (x + 2) are the factors of f(x).
So,
x – 1 = 0; x = 1 and x + 2 = 0; x = -2
f(1) and f(-2) both should be equal to zero.
(1)3 + (3m + 1)(1)2 + n(1) – 18 = 0
⇒ 1 + 3m + 1 + n – 18 = 0
⇒ 3m + n – 16 = 0 ….. (1)
And,
(-2)3 + (3m + 1)(-2)2 + n(-2) – 18 = 0
⇒ 8 + 12m + 4 – 2n – 18 = 0
⇒ 12m – 2n – 22 = 0
⇒ 6m – n – 11 = 0 ….. (2)
Adding (1) and (2), we get,
9m – 27 = 0
Thus, m = 3
Putting the value of m in (1), we have
3(3) + n – 16 =0
⇒ 9 + n – 16 = 0
∴ n = 7
Exercise 8(B)
1. Using the Factor Theorem, show that:
(i) (x – 2) is a factor of x3 – 2x2 – 9x + 18. Hence, factorise the expression x3 – 2x2 – 9x + 18 completely.
(ii) (x + 5) is a factor of 2x3 + 5x2 – 28x – 15. Hence, factorise the expression 2x3 + 5x2 – 28x – 15 completely.
(iii) (3x + 2) is a factor of 3x3 + 2x2 – 3x – 2. Hence, factorise the expression 3x3 + 2x2 – 3x – 2 completely.
Solution
(i) Here, f(x) = x3 – 2x2 – 9x + 18
So, x – 2 = 0 ⇒ x = 2
Thus, remainder = f(2)
= (2)3 – 2(2)2 – 9(2) + 18
= 8 – 8 – 18 + 18
= 0
∴ (x – 2) is a factor of f(x).
Now, performing division of polynomial f(x) by (x – 2) we have
Thus, x3 – 2x2 – 9x + 18 = (x – 2) (x2 – 9) = (x – 2) (x + 3) (x – 3)
(ii) Here, f(x) = 2x3 + 5x2 – 28x – 15
So, x + 5 = 0 ⇒ x = -5
Thus, remainder = f(-5)
= 2(-5)3 + 5(-5)2 – 28(-5) – 15
= -250 + 125 + 140 – 15
= -265 + 265
= 0
∴ (x + 5) is a factor of f(x).
Now, performing division of polynomial f(x) by (x + 5) we get
So, 2x3 + 5x2 – 28x – 15 = (x + 5) (2x2 – 5x – 3)
Further, on factorisation
= (x + 5) [2x2 – 6x + x – 3]
= (x + 5) [2x(x – 3) + 1(x – 3)] = (x + 5) (2x + 1) (x – 3)
Thus, f(x) is factorised as (x + 5) (2x + 1) (x – 3)
(iii) Here, f(x) = 3x3 + 2x2 – 3x – 2
So, 3x + 2 = 0 ⇒ x = -2/3
Thus, remainder = f(-2/3)
= 3(-2/3)3 + 2(-2/3)2 – 3(-2/3) – 2
= -8/9 + 8/9 + 2 – 2
= 0
∴ (3x + 2) is a factor of f(x).
Now, performing division of polynomial f(x) by (3x + 2) we get
Thus, 3x3 + 2x2 – 3x – 2 = (3x + 2) (x2 – 1) = (3x + 2) (x – 1) (x + 1)
2. Using the Remainder Theorem, factorise each of the following completely.
(i) 3x3 + 2x2 − 19x + 6
(ii) 2x3 + x2 – 13x + 6
(iii) 3x3 + 2x2 – 23x – 30
(iv) 4x3 + 7x2 – 36x – 63
(v) x3 + x2 – 4x – 4
Solution
(i) Let f(x) = 3x3 + 2x2 − 19x + 6
For x = 2, the value of f(x) will be
= 3(2)3 + 2(2)2 – 19(2) + 6
= 24 + 8 – 38 + 6 = 0
As f(2) = 0, so (x – 2) is a factor of f(x).
Now, performing long division we have
Thus, f(x) = (x -2) (3x2 + 8x – 3)
= (x – 2) (3x2 + 9x – x – 3)
= (x – 2) [3x(x + 3) -1(x + 3)]
= (x – 2) (x + 3) (3x – 1)
(ii) Let f(x) = 2x3 + x2 – 13x + 6
For x = 2, the value of f(x) will be
f(2) = 2(2)3 + (2)2 – 13(2) + 6 = 16 + 4 – 26 + 6 = 0
As f(2) = 0, so (x – 2) is a factor of f(x).
Now, performing long division we have
Thus, f(x) = (x -2) (2x2 + 5x – 3)
= (x – 2) [2x2 + 6x – x – 3]
= (x – 2) [2x(x + 3) -1(x + 3)]
= (x – 2) [2x(x + 3) -1(x + 3)]
= (x – 2) (2x – 1) (x + 3)
(iii) Let f(x) = 3x3 + 2x2 – 23x – 30
For x = -2, the value of f(x) will be
f(-2) = 3(-2)3 + 2(-2)2 – 23(-2) – 30
= -24 + 8 + 46 – 30 = -54 + 54 = 0
As f(-2) = 0, so (x + 2) is a factor of f(x).
Now, performing long division we have
Thus, f(x) = (x + 2) (3x2 – 4x – 15)
= (x + 2) (3x2 – 9x + 5x – 15)
= (x + 2) [3x(x – 3) + 5(x – 3)]
= (x + 2) (3x + 5) (x – 3)
(iv) Let f(x) = 4x3 + 7x2 – 36x – 63
For x = 3, the value of f(x) will be
f(3) = 4(3)3 + 7(3)2 – 36(3) – 63
= 108 + 63 – 108 – 63 = 0
As f(3) = 0, (x + 3) is a factor of f(x).
Now, performing long division we have
Thus, f(x) = (x + 3) (4x2 – 5x – 21)
= (x + 3) (4x2 – 12x + 7x – 21)
= (x + 3) [4x(x – 3) + 7(x – 3)]
= (x + 3) (4x + 7) (x – 3)
(v) Let f(x) = x3 + x2 – 4x – 4
For x = -1, the value of f(x) will be
f(-1) = (-1)3 + (-1)2 – 4(-1) – 4
= -1 + 1 + 4 – 4 = 0
As, f(-1) = 0 so (x + 1) is a factor of f(x).
Now, performing long division we have
Thus, f(x) = (x + 1) (x2 – 4)
= (x + 1) (x – 2) (x + 2)
3. Using the Remainder Theorem, factorise the expression 3x3 + 10x2 + x – 6. Hence, solve the equation 3x3 + 10x2 + x – 6 = 0.
Solution
Let’s take f(x) = 3x3 + 10x2 + x – 6
For x = -1, the value of f(x) will be
f(-1) = 3(-1)3 + 10(-1)2 + (-1) – 6 = -3 + 10 – 1 – 6 = 0
As, f(-1) = 0 so (x + 1) is a factor of f(x).
Now, performing long division we have
Thus, f(x) = (x + 1) (3x2 + 7x – 6)
= (x + 1) (3x2 + 9x – 2x – 6)
= (x + 1) [3x(x + 3) -2(x + 3)]
= (x + 1) (x + 3) (3x – 2)
Now, 3x3 + 10x2 + x – 6 = 0
(x + 1) (x + 3) (3x – 2) = 0
∴ x = -1, -3 or 2/3
4. Factorise the expression f (x) = 2x3 – 7x2 – 3x + 18. Hence, find all possible values of x for which f(x) = 0.
Solution
Let f(x) = 2x3 – 7x2 – 3x + 18
For x = 2, the value of f(x) will be
f(2) = 2(2)3 – 7(2)2 – 3(2) + 18
= 16 – 28 – 6 + 18 = 0
As f(2) = 0, (x – 2) is a factor of f(x).
Now, performing long division we have
Thus, f(x) = (x – 2) (2x2 – 3x – 9)
= (x – 2) (2x2 – 6x + 3x – 9)
= (x – 2) [2x(x – 3) + 3(x – 3)]
= (x – 2) (x – 3) (2x + 3)
Now, for f(x) = 0
(x – 2) (x – 3) (2x + 3) = 0
Hence x = 2, 3 or -3/2
5. Given that x – 2 and x + 1 are factors of f(x) = x3 + 3x2 + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).
Solution
Let f(x) = x3 + 3x2 + ax + b
As, (x – 2) is a factor of f(x), so f(2) = 0
(2)3 + 3(2)2 + a(2) + b = 0
⇒ 8 + 12 + 2a + b = 0
⇒ 2a + b + 20 = 0 …(1)
And as, (x + 1) is a factor of f(x), so f(-1) = 0
(-1)3 + 3(-1)2 + a(-1) + b = 0
⇒ -1 + 3 – a + b = 0
⇒ -a + b + 2 = 0 … (2)
Subtracting (2) from (1), we have
3a + 18 = 0
⇒ a = -6
On substituting the value of a in (ii), we have
b = a – 2 = -6 – 2 = -8
Thus, f(x) = x3 + 3x2 – 6x – 8
Now, for x = -1
f(-1) = (-1)3 + 3(-1)2 – 6(-1) – 8 = -1 + 3 + 6 – 8 = 0
∴ (x + 1) is a factor of f(x).
Now, performing long division we have
Hence, f(x) = (x + 1) (x2 + 2x – 8)
= (x + 1) (x2 + 4x – 2x – 8)
= (x + 1) [x(x + 4) – 2(x + 4)]
= (x + 1) (x + 4) (x – 2)
Exercise 8(C)
1. Show that (x – 1) is a factor of x3 – 7x2 + 14x – 8. Hence, completely factorise the given expression.
Solution
Let f(x) = x3 – 7x2 + 14x – 8
Then, for x = 1
f(1) = (1)3 – 7(1)2 + 14(1) – 8
= 1 – 7 + 14 – 8
= 0
Thus, (x – 1) is a factor of f(x).
Now, performing long division we have
Hence, f(x) = (x – 1) (x2 – 6x + 8)
= (x – 1) (x2 – 4x – 2x + 8)
= (x – 1) [x(x – 4) -2(x – 4)]
= (x – 1) (x – 4) (x – 2)
2. Using Remainder Theorem, factorise:
x3 + 10x2 – 37x + 26 completely.
Solution
Let f(x) = x3 + 10x2 – 37x + 26
From remainder theorem, we know that
For x = 1, the value of f(x) is the remainder
f(1) = (1)3 + 10(1)2 – 37(1) + 26 = 1 + 10 – 37 + 26 = 0
As f(1) = 0, x – 1 is a factor of f(x).
Now, performing long division we have
Thus, f(x) = (x – 1) (x2 + 11x – 26)
= (x – 1) (x2 + 13x – 2x – 26)
= (x – 1) [x(x + 13) – 2(x + 13)]
= (x – 1) (x + 13) (x – 2)
3. When x3 + 3x2 – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.
Solution
Let f(x) = x3 + 3x2 – mx + 4
From the question, we have
f(2) = m + 3
(2)3 + 3(2)2 – m(2) + 4 = m + 3
⇒ 8 + 12 – 2m + 4 = m + 3
⇒ 24 – 3 = m + 2m
⇒ 3m = 21
Thus, m = 7
4. What should be subtracted from 3x3 – 8x2 + 4x – 3, so that the resulting expression has x + 2 as a factor?
Solution
Let’s assume the required number to be k.
And let f(x) = 3x3 – 8x2 + 4x – 3 – k
From the question, we have
f(-2) = 0
⇒ 3(-2)3 – 8(-2)2 + 4(-2) – 3 – k = 0
⇒ -24 – 32 – 8 – 3 – k = 0
⇒ -67 – k = 0
⇒ k = -67
∴ the required number is -67.
5. If (x + 1) and (x – 2) are factors of x3 + (a + 1)x2 – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.
Solution
Let’s take f(x) = x3 + (a + 1)x2 – (b – 2)x – 6
As, (x + 1) is a factor of f(x).
Then, remainder = f(-1) = 0
(-1)3 + (a + 1)(-1)2 – (b – 2) (-1) – 6 = 0
⇒ -1 + (a + 1) + (b – 2) – 6 = 0
⇒ a + b – 8 = 0 …(1)
And as, (x – 2) is a factor of f(x).
Then, remainder = f(2) = 0
(2)3 + (a + 1) (2)2 – (b – 2) (2) – 6 = 0
⇒ 8 + 4a + 4 – 2b + 4 – 6 = 0
⇒ 4a – 2b + 10 = 0
⇒ 2a – b + 5 = 0 …(2)
Adding (1) and (2), we get
3a – 3 = 0
Thus, a = 1
Substituting the value of a in (i), we get,
1 + b – 8 = 0
Thus, b = 7
Hence, f(x) = x3 + 2x2 – 5x – 6
Now as (x + 1) and (x – 2) are factors of f(x).
So, (x + 1) (x – 2) = x2 – x – 2 is also a factor of f(x).
∴ f(x) = x3 + 2x2 – 5x – 6
= (x + 1) (x – 2) (x + 3)
6. If x – 2 is a factor of x2 + ax + b and a + b = 1, find the values of a and b.
Solution
Let f(x) = x2 + ax + b
Given, (x – 2) is a factor of f(x).
Then, remainder = f(2) = 0
(2)2 + a(2) + b = 0
⇒ 4 + 2a + b = 0
⇒ 2a + b = -4 …(1)
And also given that,
a + b = 1 …(2)
Subtracting (2) from (1), we have
a = -5
On substituting the value of a in (2), we have
b = 1 – (-5) = 6
7. Factorise x3 + 6x2 + 11x + 6 completely using factor theorem.
Solution
Let f(x) = x3 + 6x2 + 11x + 6
For x = -1, the value of f(x) is
f(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6
= -1 + 6 – 11 + 6 = 12 – 12 = 0
Thus, (x + 1) is a factor of f(x).
∴ f(x) = (x + 1) (x2 + 5x + 6)
= (x + 1) (x2 + 3x + 2x + 6)
= (x + 1) [x(x + 3) + 2(x + 3)]
= (x + 1) (x + 3) (x + 2)
8. Find the value of ‘m’, if mx3 + 2x2 – 3 and x2 – mx + 4 leave the same remainder when each is divided by x – 2.
Solution
Let f(x) = mx3 + 2x2 – 3 and g(x) = x2 – mx + 4
From the question, it’s given that f(x) and g(x) leave the same remainder when divided by (x – 2). So, we have:
f(2) = g(2)
⇒ m(2)3 + 2(2)2 – 3 = (2)2 – m(2) + 4
⇒ 8m + 8 – 3 = 4 – 2m + 4
⇒ 10m = 3
Thus, m = 3/10