# Selina Solutions Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems Exercise 8(B)

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### Access Selina Solutions Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems Exercise 8(B)

1. Using the Factor Theorem, show that:

(i) (x – 2) is a factor of x3Â – 2x2Â – 9x + 18. Hence, factorise the expression x3Â – 2x2Â – 9x + 18 completely.

(ii) (x + 5) is a factor of 2x3Â + 5x2Â – 28x – 15. Hence, factorise the expression 2x3Â + 5x2Â – 28x – 15 completely.

(iii) (3x + 2) is a factor of 3x3Â + 2x2Â – 3x – 2. Hence, factorise the expression 3x3Â + 2x2Â – 3x – 2 completely.

Solution:

(i) Here, f(x) = x3Â – 2x2Â – 9x + 18

So, x – 2 = 0Â â‡’ x = 2

Thus, remainder = f(2)

= (2)3Â – 2(2)2Â – 9(2) + 18

= 8 – 8 – 18 + 18

= 0

Therefore, (x – 2) is a factor of f(x).

Now, performing division of polynomial f(x) by (x â€“ 2) we have

Thus, x3Â – 2x2Â – 9x + 18 = (x – 2) (x2Â – 9) = (x – 2) (x + 3) (x – 3)

(ii) Here, f(x) = 2x3Â + 5x2Â – 28x – 15

So, x + 5 = 0Â â‡’ x = -5

Thus, remainder = f(-5)

= 2(-5)3Â + 5(-5)2Â – 28(-5) – 15

= -250 + 125 + 140 – 15

= -265 + 265

= 0

Therefore, (x + 5) is a factor of f(x).

Now, performing division of polynomial f(x) by (x + 5) we get

So, 2x3Â + 5x2Â – 28x – 15 = (x + 5) (2x2Â – 5x – 3)

Further, on factorisation

= (x + 5) [2x2Â – 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) = 3x3Â + 2x2Â – 3x â€“ 2

So, 3x + 2 = 0Â â‡’ x = -2/3

Thus, remainder = f(-2/3)

= 3(-2/3)3Â + 2(-2/3)2Â â€“ 3(-2/3) â€“ 2

= -8/9 + 8/9 + 2 â€“ 2

= 0

Therefore, (3x + 2) is a factor of f(x).

Now, performing division of polynomial f(x) by (3x + 2) we get

Thus, 3x3Â + 2x2Â – 3x â€“ 2 = (3x + 2) (x2 – 1) = (3x + 2) (x â€“ 1) (x + 1)

2. Using the Remainder Theorem, factorise each of the following completely.

(i)Â 3x3Â + 2x2Â âˆ’ 19x + 6

(ii) 2x3Â + x2Â – 13x + 6

(iii) 3x3Â + 2x2Â – 23x – 30

(iv) 4x3Â + 7x2Â – 36x – 63

(v) x3Â + x2Â – 4x – 4

Solution:

(i) Let f(x) = 3x3Â + 2x2Â âˆ’ 19x + 6

For x = 2, the value of f(x) will be

= 3(2)3Â + 2(2)2Â â€“ 19(2) + 6

= 24 + 8 â€“ 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) = 2x3Â + x2Â – 13x + 6

For x = 2, the value of f(x) will be

f(2) = 2(2)3Â + (2)2Â – 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 â€“ 1) (x + 3)

(iii) Let f(x) = 3x3Â + 2x2Â – 23x – 30

For x = -2, the value of f(x) will be

f(-2) = 3(-2)3Â + 2(-2)2Â – 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 â€“ 4x – 15)

= (x + 2) (3x2 â€“ 9x + 5x – 15)

= (x + 2) [3x(x â€“ 3) + 5(x â€“ 3)]

= (x + 2) (3x + 5) (x â€“ 3)

(iv) Let f(x) = 4x3Â + 7x2Â – 36x – 63

For x = 3, the value of f(x) will be

f(3) = 4(3)3Â + 7(3)2Â – 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 â€“ 5x – 21)

= (x + 3) (4x2 â€“ 12x + 7x – 21)

= (x + 3) [4x(x â€“ 3) + 7(x – 3)]

= (x + 3) (4x + 7) (x – 3)

(v) Let f(x) = x3Â + x2Â – 4x – 4

For x = -1, the value of f(x) will be

f(-1) = (-1)3Â + (-1)2Â – 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 3x3Â + 10x2Â + x – 6. Hence, solve the equation 3x3Â + 10x2Â + x – 6 = 0.

Solution:

Letâ€™s take f(x) = 3x3Â + 10x2Â + x – 6

For x = -1, the value of f(x) will be

f(-1) = 3(-1)3Â + 10(-1)2Â + (-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 â€“ 2x – 6)

= (x + 1) [3x(x + 3) -2(x + 3)]

= (x + 1) (x + 3) (3x – 2)

Now, 3x3Â + 10x2Â + x – 6 = 0

(x + 1) (x + 3) (3x – 2) = 0

Therefore,

x = -1, -3 or 2/3

4. Factorise the expression f (x) = 2x3Â – 7x2Â – 3x + 18. Hence, find all possible values of x for which f(x) = 0.

Solution:

Let f(x) = 2x3Â – 7x2Â – 3x + 18

For x = 2, the value of f(x) will be

f(2) = 2(2)3Â – 7(2)2Â – 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 â€“ 3x – 9)

= (x – 2) (2x2 â€“ 6x + 3x – 9)

= (x – 2) [2x(x â€“ 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) = x3Â + 3x2Â + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).

Solution:

Let f(x) = x3Â + 3x2Â + ax + b

As, (x – 2) is a factor of f(x), so f(2) = 0

(2)3Â + 3(2)2Â + 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)3Â + 3(-1)2Â + 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 â€“ 2 = -8

Thus, f(x) = x3Â + 3x2Â – 6x – 8

Now, for x = -1

f(-1) = (-1)3Â + 3(-1)2Â – 6(-1) – 8 = -1 + 3 + 6 – 8 = 0

Therefore, (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 â€“ 2x – 8)

= (x + 1) [x(x + 4) â€“ 2(x + 4)]

= (x + 1) (x + 4) (x – 2)