Factorization of Polynomial — Exercise 8(A)

x - 1 is a factor of 8x² - 7x + m; the value of m is :

1. -1

2. 1

3. -2

4. 2

Answer

By factor theorem,

If x - a is a factor of polynomial f(x), then remainder f(a) = 0.

Given,

x - 1 is a factor of 8x² - 7x + m.

⇒ x - 1 = 0

⇒ x = 1.

Substituting x = 1, in 8x² - 7x + m remainder will be zero.

⇒ 8(1)² - 7(1) + m = 0

⇒ 8 - 7 + m = 0

⇒ 1 + m = 0

⇒ m = -1.

Hence, Option 1 is the correct option.

If (x - a) is a factor of f(x) then the remainder when f(x) - k is divided by (x - a) is :

1. 0

2. k

3. -k

4. a

Answer

By factor theorem,

If x - a is a factor of polynomial f(x), then remainder f(a) = 0.

Given,

(x - a) is a factor of f(x), then f(a) = 0.

Substituting x = a, in f(x) - k,

⇒ f(a) - k

⇒ 0 - k

⇒ -k.

Hence, Option 3 is the correct option.

One factor of x³ - kx² + 11x - 6 is x - 1. The value of k is :

1. -6

2. 12

3. 6

4. -12

Answer

By factor theorem,

If x - a is a factor of polynomial f(x), then remainder f(a) = 0.

Given,

x - 1 is a factor of 8x² - 7x + m.

⇒ x - 1 = 0

⇒ x = 1.

Substituting x = 1, in x³ - kx² + 11x - 6 remainder will be zero.

⇒ 1³ - k(1)² + 11(1) - 6 = 0

⇒ 1 - k + 11 - 6 = 0

⇒ 6 - k = 0

⇒ k = 6.

Hence, Option 3 is the correct option.

If (x - a) is a factor of x³ - ax² + x + 5; the value of a is :

1. $\dfrac{1}{5}$

2. 5

3. -$\dfrac{1}{5}$

4. -5

Answer

By factor theorem,

If polynomial ƒ(x) is divided by its factor (x - a) then the remainder ƒ(a) = 0.

Since, x - a is a factor of x³ - ax² + x + 5.

∴ On substituting x = a in x³ - ax² + x + 5, remainder = 0.

∴ a³ - a(a)² + a + 5 = 0

⇒ a³ - a³ + a + 5 = 0

⇒ a + 5 = 0

⇒ a = -5.

Hence, Option 4 is the correct option.

(x - 2) is a factor of :

1. x³ - x² + x - 6

2. x³ + x² + x + 6

3. 2x³ - 6x² + 5x - 1

4. x³ - 4x² + x - 8

Answer

⇒ x - 2 = 0

⇒ x = 2.

Substituting x = 2 in x³ - x² + x - 6, we get :

⇒ 2³ - 2² + 2 - 6

⇒ 8 - 4 + 2 - 6

⇒ 10 - 10

⇒ 0.

Since, remainder = 0.

∴ x - 2 is a factor of x³ - x² + x - 6.

Hence, Option 1 is the correct option.

Find in each case, the remainder when :

x⁴ - 3x² + 2x + 1 is divided by x - 1

Answer

x - 1 = 0 ⇒ x = 1.

Required remainder = Value of given polynomial x⁴ - 3x² + 2x + 1 at x = 1.

∴ Remainder = (1)⁴ - 3(1)² + 2(1) + 1

= 1 - 3 + 2 + 1

= 1.

Hence, remainder = 1.

Find in each case, the remainder when :

x³ + 3x² - 12x + 4 is divided by x - 2.

Answer

x - 2 = 0 ⇒ x = 2.

Required remainder = Value of given polynomial x³ + 3x² - 12x + 4 at x = 2.

∴ Remainder = (2)³ + 3(2)² - 12(2) + 4

= 8 + 12 - 24 + 4

= 0.

Hence, remainder = 0.

Find in each case, the remainder when :

x⁴ + 1 is divisible by x + 1.

Answer

x + 1 = 0 ⇒ x = -1.

Required remainder = Value of given polynomial x⁴ + 1 at x = -1.

∴ Remainder = (-1)⁴ + 1

= 1 + 1

= 2.

Hence, remainder = 2.

Use the Remainder theorem to find which of the following is a factor of 2x³ + 3x² - 5x - 6.

(i) x + 1

(ii) 2x - 1

Answer

(i) x + 1 = 0 ⇒ x = -1

Required remainder = Value of given polynomial 2x³ + 3x² - 5x - 6 at x = -1.

∴ Remainder = 2(-1)³ + 3(-1)² - 5(-1) - 6

= 2(-1) + 3(1) + 5 - 6

= -2 + 3 + 5 - 6

= 8 - 8

Since, remainder = 0

∴ x + 1 is a factor of 2x³ + 3x² - 5x - 6

(ii) 2x - 1 = 0 ⇒ x = $\dfrac{1}{2}$

Required remainder = Value of given polynomial 2x³ + 3x² - 5x - 6 at x = $\dfrac{1}{2}$.

$\therefore \text{ Remainder} = 2\Big(\dfrac{1}{2}\Big)^3 + 3\Big(\dfrac{1}{2}\Big)^2 - 5\Big(\dfrac{1}{2}\Big) - 6 \\[1em] = 2 \times \dfrac{1}{8} + 3 \times \dfrac{1}{4} - \dfrac{5}{2} - 6 \\[1em] = \dfrac{1}{4} + \dfrac{3}{4} - \dfrac{5}{2} - 6 \\[1em] = \dfrac{1 + 3 - 10 - 24}{4} \\[1em] = -\dfrac{30}{4} \\[1em] = -\dfrac{15}{2}.$

Since, remainder ≠ 0

∴ 2x - 1 is not a factor of 2x³ + 3x² - 5x - 6.

If 2x + 1 is a factor of 2x² + ax - 3, find the value of a.

Answer

2x + 1 = 0 ⇒ x = $-\dfrac{1}{2}$

Since, 2x + 1 is a factor of 2x² + ax - 3

∴ On substituting x = $-\dfrac{1}{2}$ in 2x² + ax - 3 remainder = 0.

$\Rightarrow 2\Big(-\dfrac{1}{2}\Big)^2 + a\Big(-\dfrac{1}{2}\Big) - 3 = 0 \\[1em] \Rightarrow 2 \times \dfrac{1}{4} - \dfrac{a}{2} - 3 = 0 \\[1em] \Rightarrow \dfrac{1}{2} - \dfrac{a}{2} = 3 \\[1em] \Rightarrow \dfrac{1 - a}{2} = 3 \\[1em] \Rightarrow 1 - a = 6 \\[1em] \Rightarrow a = 1 - 6 = -5.$

Hence, a = -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.

Answer

x - 2 = 0 ⇒ x = 2

Since, x - 2 is a factor of x³ + ax² + bx - 12

∴ On substituting x = 2 in x³ + ax² + bx - 12, remainder = 0.

⇒ (2)³ + a(2)² + b(2) - 12 = 0

⇒ 8 + 4a + 2b - 12 = 0

⇒ 4a + 2b - 4 = 0

⇒ 4a + 2b = 4

⇒ 2(2a + b) = 4

⇒ 2a + b = 2

⇒ b = 2 - 2a .........(i)

x + 3 = 0 ⇒ x = -3

Since, x + 3 is a factor of x³ + ax² + bx - 12

∴ On substituting x = -3 in x³ + ax² + bx - 12, remainder = 0.

⇒ (-3)³ + a(-3)² + b(-3) - 12 = 0

⇒ -27 + 9a - 3b - 12 = 0

⇒ 9a - 3b - 39 = 0

⇒ 9a - 3b = 39

⇒ 3(3a - b) = 39

⇒ 3a - b = 13

⇒ b = 3a - 13 .........(ii)

From (i) and (ii) we get,

⇒ 2 - 2a = 3a - 13

⇒ 3a + 2a = 2 + 13

⇒ 5a = 15

⇒ a = 3.

Substituting value of a in (i),

⇒ b = 2 - 2a = 2 - 2(3) = 2 - 6 = -4.

Hence, a = 3 and b = -4.

Find the value of k, if 2x + 1 is a factor of (3k + 2)x³ + (k - 1).

Answer

2x + 1 = 0 ⇒ x = -$\dfrac{1}{2}$

Since, 2x + 1 is a factor of (3k + 2)x³ + (k - 1)

∴ On substituting x = $-\dfrac{1}{2}$ in (3k + 2)x³ + (k - 1), remainder = 0.

$\Rightarrow (3k + 2)\Big(-\dfrac{1}{2}\Big)^3 + (k - 1) = 0 \\[1em] \Rightarrow \dfrac{-(3k + 2)}{8} + (k - 1) = 0 \\[1em] \Rightarrow \dfrac{-3k - 2 + 8k - 8}{8} = 0 \\[1em] \Rightarrow 5k - 10 = 0 \\[1em] \Rightarrow 5k = 10 \\[1em] \Rightarrow k = 2.$

Hence, k = 2.

Find the values of m and n so that x - 1 and x + 2 both are factors of

x³ + (3m + 1)x² + nx - 18.

Answer

x - 1 = 0 ⇒ x = 1.

Since, x - 1 is a factor of x³ + (3m + 1)x² + nx - 18,

∴ On substituting x = 1 in x³ + (3m + 1)x² + nx - 18, remainder = 0.

⇒ (1)³ + (3m + 1)(1)² + n(1) - 18 = 0

⇒ 1 + 3m + 1 + n - 18 = 0

⇒ 3m + n - 16 = 0

⇒ n = 16 - 3m .........(i)

x + 2 = 0 ⇒ x = -2.

Since, x + 2 is a factor of x³ + (3m + 1)x² + nx - 18,

∴ On substituting x = -2 in x³ + (3m + 1)x² + nx - 18, remainder = 0.

(-2)³ + (3m + 1)(-2)² + n(-2) - 18 = 0

⇒ -8 + (3m + 1)(4) - 2n - 18 = 0

⇒ -8 + 12m + 4 - 2n - 18 = 0

⇒ 12m - 2n - 22 = 0

⇒ 12m - 2n = 22

⇒ 2(6m - n) = 22

⇒ 6m - n = 11

⇒ n = 6m - 11 .........(ii)

From (i) and (ii) we get,

⇒ 16 - 3m = 6m - 11

⇒ 6m + 3m = 16 + 11

⇒ 9m = 27

⇒ m = 3.

Substituting m = 3 in (ii) we get,

⇒ n = 6(3) - 11 = 18 - 11 = 7.

Hence, m = 3 and n = 7.

When x³ + 2x² - kx + 4 is divided by x - 2, the remainder is k. Find the value of constant k.

Answer

x - 2 = 0 ⇒ x = 2.

Given, when x³ + 2x² - kx + 4 is divided by x - 2, the remainder is k.

∴ On substituting x = 2 in x³ + 2x² - kx + 4, remainder = k.

⇒ (2)³ + 2(2)² - k(2) + 4 = k

⇒ 8 + 8 - 2k + 4 = k

⇒ 20 - 2k = k

⇒ 3k = 20

⇒ k = $\dfrac{20}{3} = 6\dfrac{2}{3}.$

Hence, k = $6\dfrac{2}{3}.$

Find the value of a, if the division of ax³ + 9x² + 4x - 10 by x + 3 leaves a remainder 5.

Answer

x + 3 = 0 ⇒ x = -3.

Given, when ax³ + 9x² + 4x - 10 is divided by x + 3, the remainder is 5.

∴ On substituting x = -3 in ax³ + 9x² + 4x - 10, remainder = 5.

⇒ a(-3)³ + 9(-3)² + 4(-3) - 10 = 5

⇒ -27a + 81 - 12 - 10 = 5

⇒ -27a + 59 = 5

⇒ 27a = 59 - 5

⇒ 27a = 54

⇒ a = $\dfrac{54}{27}$ = 2

Hence, a = 2.

If x³ + ax² + bx + 6 has x - 2 as a factor and leaves a remainder 3 when divided by x - 3, find the values of a and b.

Answer

x - 2 = 0 ⇒ x = 2.

Since, x - 2 is a factor of x³ + ax² + bx + 6,

∴ On substituting x = 2 in x³ + ax² + bx + 6, remainder = 0.

⇒ (2)³ + a(2)² + b(2) + 6 = 0

⇒ 8 + 4a + 2b + 6 = 0

⇒ 4a + 2b + 14 = 0

⇒ 2(2a + b + 7) = 0

⇒ 2a + b + 7 = 0

⇒ b = -(7 + 2a) .......(i)

x - 3 = 0 ⇒ x = 3.

Given, when x³ + ax² + bx + 6 is divided by x - 3, the remainder is 3.

∴ On substituting x = 3 in x³ + ax² + bx + 6, remainder = 3.

⇒ (3)³ + a(3)² + b(3) + 6 = 3

⇒ 27 + 9a + 3b + 6 = 3

⇒ 9a + 3b + 33 = 3

⇒ 9a + 3b = -30

⇒ 3(3a + b) = -30

⇒ 3a + b = -10

⇒ b = -10 - 3a = -(10 + 3a) ........(ii)

From (i) and (ii) we get,

⇒ -(7 + 2a) = -(10 + 3a)

⇒ 7 + 2a = 10 + 3a

⇒ 3a - 2a = 7 - 10

⇒ a = -3.

Substituting a = -3 in (i) we get,

⇒ b = -(7 + 2a) = -(7 + 2(-3)) = -(7 - 6) = -1.

Hence, a = -3 and b = -1.

What number should be added to 3x³ - 5x² + 6x so that when resulting polynomial is divided by x - 3, the remainder is 8 ?

Answer

Let number to be added be a.

∴ Polynomial = 3x³ - 5x² + 6x + a

x - 3 = 0 ⇒ x = 3

On substituting x = 3 in 3x³ - 5x² + 6x + a, remainder = 8.

∴ 3(3)³ - 5(3)² + 6(3) + a = 8

⇒ 3(27) - 5(9) + 18 + a = 8

⇒ 81 - 45 + 18 + a = 8

⇒ a + 54 = 8

⇒ a = -46

Hence, no. to be added = -46.

What number should be subtracted from x³ + 3x² - 8x + 14 so that on dividing it by x - 2, the remainder is 10 ?

Answer

Let number to be subtracted be a.

∴ Polynomial = x³ + 3x² - 8x + 14 - a

x - 2 = 0 ⇒ x = 2

On substituting x = 2 in x³ + 3x² - 8x + 14 - a, remainder = 10.

∴ (2)³ + 3(2)² - 8(2) + 14 - a = 10

⇒ 8 + 3(4) - 16 + 14 - a = 10

⇒ 8 + 12 - 16 + 14 - a = 10

⇒ 18 - a = 10

⇒ a = 18 - 10 = 8.

Hence, no. to be subtracted = 8.

The polynomial 2x³ - 7x² + ax - 6 and x³ - 8x² + (2a + 1)x - 16 leave the same remainder when divided by x - 2. Find the value of 'a'.

Answer

Given,

2x³ - 7x² + ax - 6 and x³ - 8x² + (2a + 1)x - 16 leave the same remainder when divided by x - 2.

x - 2 = 0 ⇒ x = 2

∴ On substituting x = 2 in 2x³ - 7x² + ax - 6 and x³ - 8x² + (2a + 1)x - 16 the values are equal.

∴ 2(2)³ - 7(2)² + a(2) - 6 = (2)³ - 8(2)² + (2a + 1)(2) - 16

⇒ 2(8) - 7(4) + 2a - 6 = 8 - 32 + 4a + 2 - 16

⇒ 16 - 28 + 2a - 6 = 8 - 32 + 4a + 2 - 16

⇒ 2a - 18 = 4a - 38

⇒ 4a - 2a = 38 - 18

⇒ 2a = 20

⇒ a = 10.

Hence, a = 10.

If (x - 2) is a factor of the expression 2x³ + ax² + bx - 14 and when the expression is divided by (x - 3), it leaves a remainder 52, find the values of a and b.

Answer

Given,

(x - 2) is a factor of the expression 2x³ + ax² + bx - 14.

x - 2 = 0 ⇒ x = 2

∴ On substituting x = 2 in 2x³ + ax² + bx - 14, remainder = 0.

⇒ 2(2)³ + a(2)² + b(2) - 14 = 0

⇒ 2(8) + 4a + 2b - 14 = 0

⇒ 16 + 4a + 2b - 14 = 0

⇒ 4a + 2b + 2 = 0

⇒ 2(2a + b + 1) = 0

⇒ 2a + b + 1 = 0

⇒ b = -(1 + 2a) .......(i)

Given,

On dividing 2x³ + ax² + bx - 14 by (x - 3), remainder = 52

x - 3 = 0 ⇒ x = 3

∴ On substituting x = 3 in 2x³ + ax² + bx - 14, remainder = 52.

⇒ 2(3)³ + a(3)² + b(3) - 14 = 52

⇒ 2(27) + 9a + 3b - 14 = 52

⇒ 54 + 9a + 3b - 14 = 52

⇒ 9a + 3b + 40 = 52

⇒ 9a + 3b = 12

⇒ 3(3a + b) = 12

⇒ 3a + b = 4

⇒ b = 4 - 3a ........(ii)

From (i) and (ii) we get,

⇒ -(1 + 2a) = 4 - 3a

⇒ -1 - 2a = 4 - 3a

⇒ -2a + 3a = 4 + 1

⇒ a = 5.

Substituting value of a in (i) we get,

⇒ b = -(1 + 2a) = -(1 + 2(5)) = -(1 + 10) = -11.

Hence, a = 5 and b = -11.