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 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

Therefore, (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

Therefore, (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

Therefore, (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

Therefore,

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

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 – 2x – 8)

= (x + 1) [x(x + 4) – 2(x + 4)]

= (x + 1) (x + 4) (x – 2)