Factorization of Polynomial (Remainder and Factor Theorems)

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.

Which of the following is a factor of (x - 2)² - (x² - 4) ?

1. x + 2

2. x - 2

3. x²

4. x

Answer

Given,

Polynomial = (x - 2)² - (x² - 4)

= x² + 4 - 4x - (x² - 4)

= x² - x² + 4 + 4 - 4x

= -4x + 8

On substituting x = 2 in above polynomial, we get remainder :

= -4(2) + 8 = -8 + 8 = 0.

So, (x - 2) is a factor of (x - 2)² - (x² - 4).

Hence, Option 2 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.

Show that 3x + 2 is a factor of 3x² - x - 2.

Answer

3x + 2 = 0 ⇒ x = $-\dfrac{2}{3}$

∴ When given polynomial 3x² - x - 2 is divided by x - 2, the remainder

$= 3\Big(-\dfrac{2}{3}\Big)^2 - \Big(-\dfrac{2}{3}\Big) - 2 \\[1em] = 3 \times \dfrac{4}{9} + \dfrac{2}{3} - 2 \\[1em] = \dfrac{4}{3} + \dfrac{2}{3} - 2 \\[1em] = \dfrac{6}{3} - 2 \\[1em] = 2 - 2 = 0.$

Hence, by factor theorem 3x + 2 is a factor of 3x² - x - 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

(iii) x + 2

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.

(iii) x + 2 = 0 ⇒ x = -2

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

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

= 2(-8) + 3(4) + 10 - 6

= -16 + 12 + 10 - 6

= 22 - 22

= 0.

Since, remainder = 0

∴ x + 2 is 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 value of k, if 3x - 4 is a factor of the expression 3x² + 2x - k.

Answer

3x - 4 = 0 ⇒ x = $\dfrac{4}{3}$

Since, 3x - 4 is a factor of 3x² + 2x - k

∴ On substituting x = $\dfrac{4}{3}$ in 3x² + 2x - k, remainder = 0.

$\Rightarrow 3\Big(\dfrac{4}{3}\Big)^2 + 2\Big(\dfrac{4}{3}\Big) - k = 0 \\[1em] \Rightarrow 3 \times \dfrac{16}{9} + \dfrac{8}{3} - k = 0 \\[1em] \Rightarrow \dfrac{16}{3} + \dfrac{8}{3} - k = 0 \\[1em] \Rightarrow \dfrac{16 + 8}{3} = k \\[1em] \Rightarrow k = \dfrac{24}{3} = 8.$

Hence, k = 8.

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.

For the polynomial x⁵ - x⁴ + x³ - 8x² + 6x + 15, the maximum number of linear factors is :

1. 9

2. 6

3. 7

4. 5

Answer

Maximum number of linear factors in a polynomial depends upon the highest power of variable.

Maximum number of linear factors in x⁵ - x⁴ + x³ - 8x² + 6x + 15 = 5.

Hence, Option 4 is the correct option.

If f(x) = 3x + 8; the value of f(x) + f(-x) is :

1. 8

2. 16

3. -8

4. -16

Answer

Given,

f(x) = 3x + 8

f(-x) = 3(-x) + 8 = -3x + 8

f(x) + f(-x) = 3x + 8 + (-3x + 8)

= 3x - 3x + 8 + 8

= 16.

Hence, Option 2 is the correct option.

If x²⁵ + x²⁴ is divided by (x + 1), the result is :

1. 49

2. 1

3. 0

4. -1

Answer

By remainder theorem,

The remainder theorem states that when a polynomial f(x) is divided by (x - a), then the remainder = f(a).

Given,

⇒ x + 1 = 0

⇒ x = -1.

Substituting x = -1 in x²⁵ + x²⁴, we get :

⇒ (-1)²⁵ + (-1)²⁴

⇒ -1 + 1

⇒ 0.

Hence, Option 3 is the correct option.

Factors of 3x³ - 2x² - 8x are :

1. x(3x² - 2x - 8)

2. x(x - 2)(3x + 4)

3. 2x(x - 4)(2x + 1)

4. x(x - 4)(2x + 1)

Answer

Given,

⇒ 3x³ - 2x² - 8x

⇒ x[3x² - 2x - 8]

⇒ x[3x² - 6x + 4x - 8]

⇒ x[3x(x - 2) + 4(x - 2)]

⇒ x[(3x + 4)(x - 2)]

⇒ x(3x + 4)(x - 2).

Hence, Option 2 is the correct option.

Factors of 4 + 4x - x² - x³ are :

1. (2 + x)(2 - x)(1 + x)

2. (x - 2)(1 + x)(2 + x)

3. (x + 2)(x - 2)(1 - x)

4. (2 + x)(x - 1)(2 - x)

Answer

Substituting x = 2 in 4 + 4x - x² - x³, we get :

⇒ 4 + 4x - x² - x³

⇒ 4 + 4(2) - 2² - 2³

⇒ 4 + 8 - 4 - 8

⇒ 0.

∴ (x - 2) is the factor of 4 + 4x - x² - x³.

Dividing -x³ - x² + 4x + 4 by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{-x^2 -3x - 2} \\ x - 2\overline{\smash{\big)}-x^3 - x^2 + 4x + 4} \\ \phantom{x - 2)}\phantom{2}\underline{\underset{+}{-}x^3 \underset{-}{+}2x^2} \\ \phantom{{x - 2}x^3-2}-3x^2 + 4x \\ \phantom{{x - 2)}x^3-2}\underline{\underset{+}{-}3x^2 \underset{-}{+} 6x} \\ \phantom{{x - 2)}{x^3-2x^{2}(3)}}-2x + 4 \\ \phantom{{x - 2)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}2x \underset{-}{+} 4} \\ \phantom{{x - 2)}{x^3-2x^{2}(31)}{-2x}}\times \end{array}$

we get quotient = -x² - 3x - 2.

∴ -x³ - x² + 4x + 4 = (x - 2)(-x² - 3x - 2)

= (x - 2)[-x² - 2x - x - 2]

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

= (x - 2)(x + 2)(-x - 1)

= -(x - 2)(x + 2)(x + 1)

= (2 - x)(x + 2)(x + 1)

Rearranging the terms we get,

⇒ (2 + x)(2 - x)(1 + x)

Hence, Option 1 is the correct option.

Using Factor Theorem, show that :

(x - 2) is a factor of x³ - 2x² - 9x + 18. Hence, factorise the expression x³ - 2x² - 9x + 18 completely.

Answer

x - 2 = 0 ⇒ x = 2.

Remainder = The value of x³ - 2x² - 9x + 18 at x = 2.

= (2)³ - 2(2)² - 9(2) + 18

= 8 - 8 - 18 + 18

= 0.

Hence, (x - 2) is a factor of x³ - 2x² - 9x + 18.

Now dividing x³ - 2x² - 9x + 18 by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{x^2 - 9} \\ x - 2\overline{\smash{\big)}x^3 - 2x^2 - 9x + 18} \\ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-}2x^2} \\ \phantom{{x - 2}x^3-2x^2}-9x + 18 \\ \phantom{{x - 2}x^3-2x^2\space}\underline{\underset{+}{-}9x \underset{-}{+} 18} \\ \phantom{{x - 2}{x^3-2x^2\space}{-9x}}\times \end{array}$

we get quotient = x² - 9

∴ x³ - 2x² - 9x + 18 = (x - 2)(x² - 9) = (x - 2)(x - 3)(x + 3).

Hence, x³ - 2x² - 9x + 18 = (x - 2)(x - 3)(x + 3).

Using Factor Theorem, show that :

(x + 5) is a factor of 2x³ + 5x² - 28x - 15. Hence, factorise the expression 2x³ + 5x² - 28x - 15 completely.

Answer

x + 5 = 0 ⇒ x = -5.

Remainder = The value of 2x³ + 5x² - 28x - 15 at x = -5.

= 2(-5)³ + 5(-5)² - 28(-5) - 15

= 2(-125) + 5(25) + 140 - 15

= -250 + 125 + 140 - 15

= -265 + 265

= 0.

Hence, (x + 5) is a factor of 2x³ + 5x² - 28x - 15.

Now dividing 2x³ + 5x² - 28x - 15 by (x + 5),

$\begin{array}{l} \phantom{x + 5)}{2x^2 - 5x - 3} \\ x + 5\overline{\smash{\big)}2x^3 + 5x^2 - 28x - 15} \\ \phantom{x - 5}\underline{\underset{-}{}2x^3 \underset{-}{+}10x^2} \\ \phantom{{x - 5}2x^3+}-5x^2 - 28x \\ \phantom{{x - 5}2x^3+}\underline{\underset{+}{-}5x^2 \underset{+}{-} 25x} \\ \phantom{{x - 5}{2x^3+}{+5x^2+}}-3x - 15 \\ \phantom{{x - 5}{2x^3+}{+5x^2+\enspace}}\underline{\underset{+}{-}3x \underset{+}{-} 15} \\ \phantom{{x - 5}{2x^3+}{+5x^2+\enspace}{-3x}}\times \end{array}$

we get quotient = 2x² - 5x - 3

Factorising 2x² - 5x - 3,

⇒ 2x² - 6x + x - 3

⇒ 2x(x - 3) + 1(x - 3)

⇒ (2x + 1)(x - 3)

∴ 2x³ + 5x² - 28x - 15 = (x + 5)(2x + 1)(x - 3).

Hence, 2x³ + 5x² - 28x - 15 = (x + 5)(2x + 1)(x - 3).

Using the remainder theorem, factorise each of the following completely :

3x³ + 2x² - 19x + 6

Answer

For x = 2, the value of 3x³ + 2x² - 19x + 6,

= 3(2)³ + 2(2)² - 19(2) + 6

= 3(8) + 2(4) - 38 + 6

= 38 - 38

= 0.

Hence, (x - 2) is the factor of 3x³ + 2x² - 19x + 6.

On dividing 3x³ + 2x² - 19x + 6 by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{3x^2 + 8x - 3} \\ x - 2\overline{\smash{\big)}3x^3 + 2x^2 - 19x + 6} \\ \phantom{x - 2}\underline{\underset{-}{}3x^3 \underset{+}{-} 6x^2} \\ \phantom{{x - 2}2x^3+4}8x^2 - 19x \\ \phantom{{x - 2}2x^3+}\underline{\underset{-}{}8x^2 \underset{-}{+} 16x} \\ \phantom{{x - 2}{2x^3+}{+2x^2}}-3x + 6 \\ \phantom{{x - 2}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}3x \underset{-}{+} 6} \\ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, quotient = 3x² + 8x - 3

Factorising, 3x² + 8x - 3

= 3x² + 9x - x - 3

= 3x(x + 3) - 1(x + 3)

= (3x - 1)(x + 3)

∴ 3x² + 8x - 3 = (3x - 1)(x + 3).

Hence, 3x³ + 2x² - 19x + 6 = (x - 2)(3x - 1)(x + 3).

Using the remainder theorem, factorise each of the following completely :

2x³ + x² - 13x + 6

Answer

For x = 2, the value of 2x³ + x² - 13x + 6

= 2(2)³ + (2)² - 13(2) + 6

= 2(8) + 4 - 26 + 6

= 26 - 26

= 0.

On dividing 2x³ + x² - 13x + 6 by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{2x^2 + 5x - 3} \\ x - 2\overline{\smash{\big)}2x^3 + x^2 - 13x + 6} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-} 4x^2} \\ \phantom{{x - 2}2x^3+4}5x^2 - 13x \\ \phantom{{x - 2}2x^3+}\underline{\underset{-}{}5x^2 \underset{+}{-} 10x} \\ \phantom{{x - 2}{2x^3+}{+2x^2}}-3x + 6 \\ \phantom{{x - 2}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}3x \underset{-}{+} 6} \\ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, quotient = 2x² + 5x - 3

Factorising, 2x² + 5x - 3

= 2x² + 6x - x - 3

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

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

∴ 2x² + 5x - 3 = (2x - 1)(x + 3)

Hence, 2x³ + x² - 13x + 6 = (x - 2)(2x - 1)(x + 3).

Using the remainder theorem, factorise each of the following completely :

3x³ + 2x² - 23x - 30

Answer

For x = -2 the value of 3x³ + 2x² - 23x - 30,

= 3(-2)³ + 2(-2)² - 23(-2) - 30

= 3(-8) + 2(4) + 46 - 30

= -24 + 8 + 46 - 30

= 54 - 54

= 0.

Hence, (x + 2) is the factor of 3x³ + 2x² - 23x - 30.

On dividing 3x³ + 2x² - 23x - 30 by x + 2,

$\begin{array}{l} \phantom{x + 2)}{3x^2 - 4x - 15} \\ x + 2\overline{\smash{\big)}3x^3 + 2x^2 - 23x - 30} \\ \phantom{x + 2}\underline{\underset{-}{}3x^3 \underset{-}{+} 6x^2} \\ \phantom{{x + 2}2x^3+}-4x^2 - 23x \\ \phantom{{x + 2}2x^3+}\underline{\underset{+}{-}4x^2 \underset{+}{-} 8x} \\ \phantom{{x + 2}{2x^3+}{+2x^2}}-15x - 30 \\ \phantom{{x + 2}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}15x \underset{+}{-} 30} \\ \phantom{{x + 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, quotient = 3x² - 4x - 15

Factorising, 3x² - 4x - 15

= 3x² - 9x + 5x - 15

= 3x(x- 3) + 5(x - 3)

= (3x + 5)(x - 3).

∴ 3x² - 4x - 15 = (3x + 5)(x - 3).

Hence, 3x³ + 2x² - 23x - 30 = (x + 2)(3x + 5)(x - 3).

Using the remainder theorem, factorise each of the following completely :

4x³ + 7x² - 36x - 63

Answer

For x = -3 the value of 4x³ + 7x² - 36x - 63

= 4(-3)³ + 7(-3)² - 36(-3) - 63

= 4(-27) + 7(9) + 108 - 63

= -108 + 63 + 108 - 63

= 0.

Hence, (x + 3) is the factor of 4x³ + 7x² - 36x - 63.

On dividing 4x³ + 7x² - 36x - 63 by (x + 3),

$\begin{array}{l} \phantom{x + 3)}{4x^2 - 5x - 21} \\ x + 3\overline{\smash{\big)}4x^3 + 7x^2 - 36x - 63} \\ \phantom{x + 3}\underline{\underset{-}{}4x^3 \underset{-}{+} 12x^2} \\ \phantom{{x + 3}2x^3+}-5x^2 - 36x \\ \phantom{{x + 3}2x^3+}\underline{\underset{+}{-}5x^2 \underset{+}{-} 15x} \\ \phantom{{x + 3}{2x^3+}{+2x^2}}-21x - 63 \\ \phantom{{x + 3}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}21x \underset{+}{-} 63} \\ \phantom{{x + 3}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, quotient = 4x² - 5x - 21

Factorising 4x² - 5x - 21,

= 4x² - 12x + 7x - 21

= 4x(x - 3) + 7(x - 3)

= (4x + 7)(x - 3).

∴ 4x² - 5x - 21 = (4x + 7)(x - 3)

Hence, 4x³ + 7x² - 36x - 63 = (x + 3)(4x + 7)(x - 3).

Using the remainder theorem, factorise each of the following completely :

x³ + x² - 4x - 4

Answer

For x = -1 the value of x³ + x² - 4x - 4

= (-1)³ + (-1)² - 4(-1) - 4

= -1 + 1 + 4 - 4

= 0.

Hence, (x + 1) is the factor of x³ + x² - 4x - 4.

On dividing x³ + x² - 4x - 4 by (x + 1),

$\begin{array}{l} \phantom{x + 1)}{x^2 - 4} \\ x + 1\overline{\smash{\big)}x^3 + x^2 - 4x - 4} \\ \phantom{x + 1}\underline{\underset{-}{}x^3 \underset{-}{+} x^2} \\ \phantom{{x + 1}x^3+x^2-}-4x - 4 \\ \phantom{{x + 1}x^3+x^2-}\underline{\underset{+}{-}4x \underset{+}{-} 4} \\ \phantom{{x + 1}x^3+x^2-4x\enspace} \times \end{array}$

we get, quotient = x² - 4

Factorising x² - 4,

= (x)² - 4

= (x + 2)(x - 2)

∴ x² - 4 = (x - 2)(x + 2)

Hence, x³ + x² - 4x - 4 = (x + 1)(x + 2)(x - 2).

Using the Remainder Theorem, factorise the expression 3x³ + 10x² + x - 6. Hence, solve the equation 3x³ + 10x² + x - 6 = 0.

Answer

For x = -1, the value of 3x³ + 10x² + x - 6,

= 3(-1)³ + 10(-1)² + (-1) - 6

= 3(-1) + 10(1) - 7

= 10 - 10

= 0.

Hence, (x + 1) is the factor of 3x³ + 10x² + x - 6.

On dividing 3x³ + 10x² + x - 6 by (x + 1),

$\begin{array}{l} \phantom{x + 1)}{3x^2 + 7x - 6} \\ x + 1\overline{\smash{\big)}3x^3 + 10x^2 + x - 6} \\ \phantom{x + 1}\underline{\underset{-}{}3x^3 \underset{-}{+} 3x^2} \\ \phantom{{x + 1}3x^3+10}7x^2 + x \\ \phantom{{x + 1}3x^3+1}\underline{\underset{-}{}7x^2 \underset{-}{+} 7x} \\ \phantom{{x + 1}{3x^3+1}{+2x^2}}-6x - 6 \\ \phantom{{x + 1}{2x^3+}{+2x^2}{41\space}}\underline{\underset{+}{-}6x \underset{+}{-} 6} \\ \phantom{{x + 1}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, quotient = 3x² + 7x - 6

Factorising 3x² + 7x - 6,

= 3x² + 9x - 2x - 6

= 3x(x + 3) - 2(x + 3)

= (3x - 2)(x + 3).

∴ 3x² + 7x - 6 = (3x - 2)(x + 3).

Hence, 3x³ + 10x² + x - 6 = (x + 1)(3x - 2)(x + 3).

Factorise the expression

f(x) = 2x³ - 7x² - 3x + 18.

Hence, find all possible values of x for which f(x) = 0.

Answer

For x = 2, the value of 2x³ - 7x² - 3x + 18,

= 2(2)³ - 7(2)² - 3(2) + 18

= 16 - 28 - 6 + 18

= 34 - 34

= 0.

Hence, (x - 2) is the factor of 2x³ - 7x² - 3x + 18.

On dividing, 2x³ - 7x² - 3x + 18 by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{2x^2 - 3x - 9} \\ x - 2\overline{\smash{\big)}2x^3 - 7x^2 - 3x + 18} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-} 4x^2} \\ \phantom{{x - 2}3x^3+}-3x^2 - 3x \\ \phantom{{x - 2}3x^3+}\underline{\underset{+}{-}3x^2 \underset{-}{+} 6x} \\ \phantom{{x - 2}{3x^3+1}{+2x^2}}-9x + 18 \\ \phantom{{x - 2}{2x^3+}{+2x^2}{41\space}}\underline{\underset{+}{-}9x \underset{-}{+} 18} \\ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get quotient = 2x² - 3x - 9.

Factorising 2x² - 3x - 9,

= 2x² - 6x + 3x - 9

= 2x(x - 3) + 3(x - 3)

= (2x + 3)(x - 3).

∴ 2x² - 3x - 9 = (2x + 3)(x - 3).

∴ f(x) = 2x³ - 7x² - 3x + 18 = (x - 2)(2x + 3)(x - 3).

f(x) = 0, if (x - 2) = 0, (2x + 3) = 0 or x - 3 = 0.

x - 2 = 0 ⇒ x =2,

2x + 3 = 0 ⇒ x = -$\dfrac{3}{2}$,

x - 3 = 0 ⇒ x = 3.

Hence, 2x³ - 7x² - 3x + 18 = (x - 2)(2x + 3)(x - 3), values for which f(x) = 0 are 2, 3, $-\dfrac{3}{2}$.

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).

Answer

x - 2 = 0 ⇒ x = 2.

Since, x - 2 is a factor of x³ + 3x² + ax + b. Hence, on substituting x = 2 in above expression, remainder = 0.

⇒ (2)³ + 3(2)² + a(2) + b = 0

⇒ 8 + 12 + 2a + b = 0

⇒ 2a + b = -20

⇒ b = -20 - 2a ........(i)

x + 1 = 0 ⇒ x = -1.

Since, x + 1 is a factor of x³ + 3x² + ax + b. Hence, on substituting x = -1 in above expression, remainder = 0.

⇒ (-1)³ + 3(-1)² + a(-1) + b = 0

⇒ -1 + 3 - a + b = 0

⇒ 2 - a + b = 0

⇒ b = a - 2 .......(ii)

From (i) and (ii) we get,

⇒ -20 - 2a = a - 2

⇒ a + 2a = -20 + 2

⇒ 3a = -18

⇒ a = -6.

Substituting value of a in (ii) we get,

⇒ b = a - 2 = -6 - 2 = -8.

∴ a = -6 and b = -8.

On dividing, x³ + 3x² - 6x - 8 by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{x^2 + 5x + 4} \\ x - 2\overline{\smash{\big)}x^3 + 3x^2 - 6x - 8} \\ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-} 2x^2} \\ \phantom{{x - 2}3x^3+}5x^2 - 6x \\ \phantom{{x - 2}3x^3\enspace\space}\underline{\underset{-}{}5x^2 \underset{+}{-} 10x} \\ \phantom{{x - 2}{3x^3+1}{+2x^2}}4x - 8 \\ \phantom{{x - 2}{2x^3+}{+2x^2}{4}}\underline{\underset{-}{}4x \underset{+}{-} 8} \\ \phantom{{x - 2}{2x^3+}{+2x^2-}{4x}}\times \end{array}$

we get, quotient = x² + 5x + 4.

Factorising x² + 5x + 4,

= x² + 4x + x + 4

= x(x + 4) + 1(x + 4)

= (x + 1)(x + 4).

Hence, a = -6, b = -8 and f(x) = (x - 2)(x + 1)(x + 4).

The expression 4x³ - bx² + x - c leaves remainders 0 and 30 when divided by x + 1 and 2x - 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.

Answer

x + 1 = 0 ⇒ x = -1.

Since, x + 1 is a factor of 4x³ - bx² + x - c. Hence, on substituting x = -1 in above expression, remainder = 0.

⇒ 4(-1)³ - b(-1)² + (-1) - c = 0

⇒ -4 - b - 1 - c = 0

⇒ c = -5 - b .......(i)

Given, on dividing 4x³ - bx² + x - c by (2x - 3) we get remainder = 30.

∴ On substituting x = $\dfrac{3}{2}$ in 4x³ - bx² + x - c, value = 30.

$\Rightarrow 4\Big(\dfrac{3}{2}\Big)^3 - b\Big(\dfrac{3}{2}\Big)^2 + \Big(\dfrac{3}{2}\Big) - c = 30 \\[1em] \Rightarrow 4\Big(\dfrac{27}{8}\Big) - b\Big(\dfrac{9}{4}\Big) + \Big(\dfrac{3}{2}\Big) - c = 30 \\[1em] \Rightarrow \dfrac{27}{2} - \dfrac{9b}{4} + \dfrac{3}{2} - 30 = c \\[1em] \Rightarrow \dfrac{54 - 9b + 6 - 120}{4} = c \\[1em] \Rightarrow \dfrac{-9b - 60}{4} = c ........(ii)$

From (i) and (ii) we get,

$\Rightarrow -5 - b = \dfrac{-9b - 60}{4} \\[1em] \Rightarrow -20 - 4b = -9b - 60 \\[1em] \Rightarrow -20 + 60 = -9b + 4b \\[1em] \Rightarrow -5b = 40 \\[1em] \Rightarrow b = -8.$

Substituting value of b = -8 in (i) we get,

c = -5 - b = -5 - (-8) = 3.

Substituting b = -8 and c = 3 in 4x³ - bx² + x - c we get,

Expression = 4x³ + 8x² + x - 3

Dividing, 4x³ + 8x² + x - 3 by (x + 1),

$\begin{array}{l} \phantom{x + 1)}{4x^2 + 4x - 3} \\ x + 1\overline{\smash{\big)}4x^3 + 8x^2 + x - 3} \\ \phantom{x + 1}\underline{\underset{-}{}4x^3 \underset{-}{+} 4x^2} \\ \phantom{{x + 1}3x^3+}4x^2 + x \\ \phantom{{x + 1}3x^3\enspace\space}\underline{\underset{-}{}4x^2 \underset{-}{+} 4x} \\ \phantom{{x + 1}{3x^3+1}{+2}}-3x - 3 \\ \phantom{{x + 1}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}3x \underset{+}{-} 3} \\ \phantom{{x + 1}{2x^3+}{+2x^2-}{4x}}\times \end{array}$

we get, quotient = 4x² + 4x - 3.

Factorising 4x² + 4x - 3,

= 4x² + 6x - 2x - 3

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

= (2x - 1)(2x + 3).

Hence, b = -8, c = 3 and x³ + 8x² + x - 3 = (x + 1)(2x - 1)(2x + 3).

If x + a is a common factor of expressions f(x) = x² + px + q and g(x) = x² + mx + n; show that : a = $\dfrac{n - q}{m - p}$

Answer

x + a = 0 ⇒ x = -a.

Since, (x + a) is factor of f(x) and g(x).

∴ f(-a) = g(-a)

⇒ (-a)² + p(-a) + q = (-a)² + m(-a) + n

⇒ a² - pa + q = a² - ma + n

⇒ a² - a² - pa + ma = n - q

⇒ ma - pa = n - q

⇒ a(m - p) = n - q

⇒ a = $\dfrac{n - q}{m - p}$.

Hence, proved that a = $\dfrac{n - q}{m - p}$.

The remainder, when x³ - x² + x - 1 is divided by x + 1, is :

1. 0

2. -4

3. 2

4. 4

Answer

By remainder theorem,

The remainder theorem states that when a polynomial f(x) is divided by (x - a), then the remainder = f(a).

Given,

⇒ x + 1 = 0

⇒ x = -1.

Substituting x = -1 in x³ - x² + x - 1 , we get :

⇒ (-1)³ - (-1)² + (-1) - 1

⇒ -1 - 1 - 1 - 1

⇒ -4.

Hence, Option 2 is the correct option.

If x - 3 is a factor of x² + kx + 15; the value of k is :

1. 8

2. 3

3. -5

4. -8

Answer

By factor theorem,

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

Given,

x - 3 is a factor of x² + kx + 15.

Then substituting x = 3 in polynomial we get, remainder = 0 :

⇒ 3² + 3k + 15 = 0

⇒ 9 + 3k + 15 = 0

⇒ 3k + 24 = 0

⇒ 3k = -24

⇒ k = $-\dfrac{24}{3}$ = -8.

Hence, Option 4 is the correct option.

Is (x - 2) a factor of x³ - 4x² - 11x + 30 ?

1. yes

2. no

3. nothing can be said

4. none of the above is true

Answer

Given,

⇒ x - 2 = 0

⇒ x = 2.

Substituting x = 2 in x³ - 4x² - 11x + 30, we get :

⇒ 2³ - 4(2)² - 11(2) + 30

⇒ 8 - 4(4) - 22 + 30

⇒ 8 - 16 - 22 + 30

⇒ 38 - 38

⇒ 0.

Since, remainder = 0.

∴ x - 2 is factor of x³ - 4x² - 11x + 30.

Hence, Option 1 is the correct option.

4x² - kx + 5 leaves a remainder 2 when divided by x - 1. The value of k is :

1. -6

2. 6

3. 7

4. -7

Answer

By remainder theorem,

If polynomial ƒ(x) is divided by (x - a) then the remainder will be ƒ(a).

Given,

4x² - kx + 5 leaves a remainder 2 when divided by x - 1.

∴ 4(1)² - k(1) + 5 = 2

⇒ 4(1) - k + 5 = 2

⇒ 4 - k + 5 = 2

⇒ 9 - k = 2

⇒ k = 9 - 2 = 7.

Hence, Option 3 is the correct option.

If mx² - nx + 8 has x - 2 as a factor, then :

1. 2m - n = 4

2. 2m + n = 4

3. 2n + m = 4

4. n - 2m = 4

Answer

By factor theorem,

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

Given,

x - 2 is a factor of mx² - nx + 8.

Then substituting x = 2 in polynomial we get, remainder = 0 :

⇒ m(2)² - 2n + 8 = 0

⇒ 4m - 2n + 8 = 0

⇒ 2(2m - n + 4) = 0

⇒ 2m - n + 4 = 0

⇒ n - 2m = 4.

Hence, Option 4 is the correct option.

Two polynomials x³⁶ - 3x³⁵ and x - 3.

Assertion (A) : If x - 3 is a factor of x³⁶ - 3x³⁵, the remainder is zero.

Reason (R) : The polynomial x - a is factor of polynomial p(x) = x³⁶ - ax³⁵, if p(a) = 0

options

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Both A and R are true and R is correct reason for A.

Reason

By factor theorem,

(x - a) is a factor of the polynomial f(x), if the remainder i.e. f(a) = 0.

Let f(x) = x³⁶ - 3x³⁵

When, x - 3 is a factor of f(x), then f(3) = 0, by factor theorem.

∴ Assertion (A) is true.

When, p(x) = x³⁶ - ax³⁵ is divided by x - a, we get :

Remainder, p(a) = a³⁶ - a.a³⁵

= a³⁶ - a³⁶

= 0.

Since, p(a) = 0, thus (x - a) is factor of p(x).

∴ Reason (R) is true.

Thus, Both A and R are true and R is correct reason for R.

Hence, option 3 is the correct option.

The polynomial 3x³ + 8x² - 15x + k and one of its factors as (x - 1).

Assertion (A) : The value of k = 4.

Reason (R) : x - 1 = 0 ⇒ x = 1.

∴ 3.(1)³ + 8.(1)² - 15 x (1) + k = 0

options

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Both A and R are true and R is correct reason for A.

Reason

Let, f(x) = 3x³ + 8x² - 15x + k

By factor theorem,

(x - a) is a factor of the polynomial f(x), if the remainder i.e. f(a) = 0.

⇒ x - 1 = 0

⇒ x = 1.

Given,

x - 1 is one of the factors of f(x).

∴ f(1) = 0

⇒ 3.(1)³ + 8.(1)² - 15.1 + k = 0

⇒ 3.1 + 8.1 - 15 + k = 0

⇒ 3 + 8 - 15 + k = 0

⇒ -4 + k = 0

⇒ k = 4.

Thus, both A and R are true and R is correct reason for A.

Hence, Option 3 is the correct option.

The polynomial x² + x + b has (x + 3) as a factor of it.

Statement 1: The value of b is -4.

Statement 2: (x + 3) is a factor of x² + x + b ⇒ (3)² + 3 + b = 0.

option

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Both the statements are false.

Reason

By factor theorem,

(x - a) is a factor of the polynomial f(x), if the remainder i.e. f(a) = 0.

⇒ x + 3 = 0

⇒ x = -3

Let, f(x) = x² + x + b

Given,

The polynomial x² + x + b has (x + 3) as a factor of it.

⇒ (-3)² + (-3) + b = 0

⇒ 9 - 3 + b = 0

⇒ 6 + b = 0

⇒ b = -6.

∴ Statement 1 is incorrect.

Since,

⇒ (-3)² + (-3) + b = 0

∴ Statement 2 is incorrect.

Hence, option 2 is the correct option.

A polynomial x⁴ - 13x² + 36.

Statement 1: x - 2 is a factor of x⁴ - 13x² + 36.

Statement 2: (2)⁴ - 13 x (2)² + 36 = 0.

option

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Both the statements are true.

Reason

By factor theorem,

(x - a) is a factor of the polynomial f(x), if the remainder i.e. f(a) = 0.

Let, f(x) = x⁴ - 13x² + 36

⇒ f(2) = 2⁴ - 13 x 2² + 36

= 16 - 52 + 36

= 0

Since, f(2) = 0,

So, x - 2 is factor of x⁴ - 13x² + 36.

∴ Statement 1 is correct.

Also,

⇒ 2⁴ - 13 x 2² + 36 = 0.

∴ Statement 2 is correct.

Hence, option 1 is the correct option.

When x³ + 3x² - mx + 4 is divided by x - 2, the remainder is m + 3. Find the value of m.

Answer

Given,

When x³ + 3x² - mx + 4 is divided by x - 2, the remainder is m + 3.

∴ (2)³ + 3(2)² - m(2) + 4 = m + 3

⇒ 8 + 12 - 2m + 4 = m + 3

⇒ 24 - 2m = m + 3

⇒ m + 2m = 24 - 3

⇒ 3m = 21

⇒ m = 7.

Hence, m = 7.

What should be subtracted from 3x³ - 8x² + 4x - 3, so that the resulting expression has x + 2 as a factor ?

Answer

The number to be subtracted = Remainder obtained on dividing 3x³ - 8x² + 4x - 3 by x + 2.

x + 2 = 0 ⇒ x = -2.

Substituting x = -2 in 3x³ - 8x² + 4x - 3 we get,

= 3(-2)³ - 8(-2)² + 4(-2) - 3

= 3(-8) - 8(4) - 8 - 3

= -24 - 32 - 8 - 3

= -67.

Hence, no. to be subtracted = -67.

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.

Answer

Given,

(x + 1) and (x - 2) are factors of x³ + (a + 1)x² - (b - 2)x - 6

x + 1 = 0 ⇒ x = -1.

Since, x + 1 is factor of x³ + (a + 1)x² - (b - 2)x - 6. Hence, on substituting x = -1 in expression, remainder = 0.

⇒ (-1)³ + (a + 1)(-1)² - (b - 2)(-1) - 6 = 0

-1 + (a + 1)(1) - (-b + 2) - 6 = 0

-1 + a + 1 + b - 2 - 6 = 0

a + b = 8

b = 8 - a ........(i)

x - 2 = 0 ⇒ x = 2.

Since, x - 2 is factor of x³ + (a + 1)x² - (b - 2)x - 6. Hence, on substituting x = 2 in expression, remainder = 0.

⇒ (2)³ + (a + 1)(2)² - (b - 2)(2) - 6 = 0

⇒ 8 + (a + 1)(4) - (2b - 4) - 6 = 0

⇒ 8 + 4a + 4 - 2b + 4 - 6 = 0

⇒ 10 + 4a - 2b = 0

⇒ 2b = 4a + 10

⇒ b = 2a + 5 ........(ii)

From (i) and (ii) we get,

⇒ 8 - a = 2a + 5

⇒ 2a + a = 8 - 5

⇒ 3a = 3.

⇒ a = 1.

Substituting a = 1, in (i) we get,

⇒ b = 8 - 1 = 7.

Substituting a = 1, b = 7 in expression we get,

Expression = x³ + (1 + 1)x² - (7 - 2)x - 6 = x³ + 2x² - 5x - 6.

On dividing, x³ + 2x² - 5x - 6 by (x + 1),

$\begin{array}{l} \phantom{x + 1)}{x^2 + x - 6} \\ x + 1\overline{\smash{\big)}x^3 + 2x^2 - 5x - 6} \\ \phantom{x + 1}\underline{\underset{-}{}x^3 \underset{+}{-} x^2} \\ \phantom{{x + 1}3x^3+}x^2 - 5x \\ \phantom{{x + 1}3x^3\enspace}\underline{\underset{-}{}x^2 \underset{-}{+} x} \\ \phantom{{x + 1}{x^32x^2}{+2}}-6x - 6 \\ \phantom{{x + 1}{x^32x^2}{+2x}}\underline{\underset{+}{-}6x \underset{+}{-} 6} \\ \phantom{{x + 1}{x^32x^2}{+2x^2-}}\times \end{array}$

we get, quotient = x² + x - 6.

Factorising, x² + x - 6

= x² + 3x - 2x - 6

= x(x + 3) - 2(x + 3)

= (x - 2)(x + 3).

Hence, a = 1, b = 7 and x³ + 2x² - 5x - 6 = (x + 1)(x - 2)(x + 3).

If x - 2 is a factor of x² + ax + b and a + b = 1, find the values of a and b.

Answer

Given,

x - 2 is a factor of x² + ax + b.

Hence, on substituting x = 2 in x² + ax + b, remainder = 0.

⇒ (2)² + a(2) + b = 0

⇒ 4 + 2a + b = 0

⇒ b = -(2a + 4)

Substituting value of b in a + b = 1 we get,

⇒ a + -(2a + 4) = 1

⇒ a - 2a - 4 = 1

⇒ -a - 4 = 1

⇒ -a = 1 + 4

⇒ a = -5.

⇒ b = -(2a + 4) = -[2(-5) + 4] = -(-10 + 4) = 6.

Hence, a = -5 and b = 6.

Find the value of 'm', if mx³ + 2x² - 3 and x² - mx + 4 leave the same remainder when divided by x - 2.

Answer

x - 2 = 0 ⇒ x = 2.

Given,

mx³ + 2x² - 3 and x² - mx + 4 leave the same remainder when divided by x - 2.

∴ m(2)³ + 2(2)² - 3 = (2)² - m(2) + 4

8m + 8 - 3 = 4 - 2m + 4

8m + 2m = 8 - 8 + 3

10m = 3

m = $\dfrac{3}{10}$.

Hence, m = $\dfrac{3}{10}$.

The polynomial px³ + 4x² - 3x + q is completely divisible by x² - 1; find the values of p and q. Also for these values of p and q factorize the given polynomial completely.

Answer

x² - 1 is a factor of px³ + 4x² - 3x + q.

∴ (x - 1) and (x + 1) are factors of px³ + 4x² - 3x + q.

Hence, substituting x = 1, -1 remainder = 0..

⇒ p(1)³ + 4(1)² - 3(1) + q = 0

⇒ p + 4 - 3 + q = 0

⇒ p + q = -1

⇒ p = -1 - q .......(i)

p(-1)³ + 4(-1)² - 3(-1) + q = 0

⇒ -p + 4 + 3 + q = 0

⇒ p = 7 + q .......(ii)

From (i) and (ii) we get,

⇒ -1 - q = 7 + q

⇒ 2q = -1 - 7

⇒ 2q = -8

⇒ q = -4.

Substituting q = -4 in (i) we get,

⇒ p = -1 - (-4) = -1 + 4 = 3.

Substituting p = 3 and q = -4 in px³ + 4x² - 3x + q,

= 3x³ + 4x² - 3x - 4.

On dividing, 3x³ + 4x² - 3x - 4 by x - 1,

$\begin{array}{l} \phantom{x - 1)}{3x^2 + 7x + 4} \\ x - 1\overline{\smash{\big)}3x^3 + 4x^2 - 3x - 4} \\ \phantom{x - 1}\underline{\underset{-}{}3x^3 \underset{+}{-} 3x^2} \\ \phantom{{x - 1}3x^3+3}7x^2 - 3x \\ \phantom{{x - 1}3x^3+}\underline{\underset{-}{}7x^2 \underset{+}{-} 7x} \\ \phantom{{x - 1}{3x^3+3x^2+}{+2}}4x - 4 \\ \phantom{{x - 1}{3x^3+3x^2+}\enspace \space}\underline{\underset{-}{}4x \underset{+}{-} 4} \\ \phantom{{x - 1}{3x^3+3x^2+}{+2-}}\times \end{array}$

we get, quotient = 3x² + 7x + 4

Factorising 3x² + 7x + 4,

= 3x² + 3x + 4x + 4

= 3x(x + 1) + 4(x + 1)

= (3x + 4)(x + 1).

Hence, p = 3, q = -4 and 3x³ + 4x² - 3x - 4 = (x - 1)(x + 1)(3x + 4).

When the polynomial x³ + 2x² - 5ax - 7 is divided by (x - 1), the remainder is A and when the polynomial x³ + ax² - 12x + 16 is divided by (x + 2), the remainder is B. Find the value of 'a' if 2A + B = 0.

Answer

Given,

When the polynomial x³ + 2x² - 5ax - 7 is divided by (x - 1), remainder is A.

x - 1 = 0 ⇒ x = 1

Substituting x = 1 in x³ + 2x² - 5ax - 7 will give, remainder = A.

∴ (1)³ + 2(1)² - 5a(1) - 7 = A

⇒ 1 + 2 - 5a - 7 = A

⇒ A = -(4 + 5a) ......(i)

Given,

When the polynomial x³ + ax² - 12x + 16 is divided by (x + 2), remainder is B.

∴ x³ + ax² - 12x + 16 = B

⇒ (-2)³ + a(-2)² - 12(-2) + 16 = B

⇒ -8 + 4a + 24 + 16 = B

⇒ B = 32 + 4a ......(ii)

Given, 2A + B = 0

∴ -2(4 + 5a) + 32 + 4a = 0

⇒ -8 - 10a + 32 + 4a = 0

⇒ 24 - 6a = 0

⇒ 6a = 24

⇒ a = 4.

Hence, a = 4.

(3x + 5) is a factor of the polynomial (a - 1)x³ + (a + 1)x² - (2a + 1)x - 15. Find the value of 'a'. For this value of 'a', factorise the given polynomial completely.

Answer

3x + 5 = 0 ⇒ x = -$\dfrac{5}{3}$

Since, (3x + 5) is a factor of the polynomial (a - 1)x³ + (a + 1)x² - (2a + 1)x - 15. Substituting x = -$\dfrac{5}{3}$ in (a - 1)x³ + (a + 1)x² - (2a + 1)x - 15, remainder = 0.

$\Rightarrow (a - 1)\Big(-\dfrac{5}{3}\Big)^3 + (a + 1)\Big(-\dfrac{5}{3}\Big)^2 - (2a + 1)\Big(-\dfrac{5}{3}\Big) - 15 = 0 \\[1em] \Rightarrow (a - 1)\Big(-\dfrac{125}{27}\Big) + (a + 1)\Big(\dfrac{25}{9}\Big) + \dfrac{10a + 5}{3} - 15 = 0 \\[1em] \Rightarrow \dfrac{-125a + 125}{27} + \dfrac{25a + 25}{9} + \dfrac{10a + 5}{3} = 15 \\[1em] \Rightarrow \dfrac{-125a + 125 + 75a + 75 + 90a + 45}{27} = 15 \\[1em] \Rightarrow 40a + 245 = 405 \\[1em] \Rightarrow 40a = 160 \\[1em] \Rightarrow a = 4.$

Substituting a = 4 in (a - 1)x³ + (a + 1)x² - (2a + 1)x - 15,

⇒ (4 - 1)x³ + (4 + 1)x² - (2(4) + 1)x - 15

⇒ 3x³ + 5x² - 9x - 15

⇒ x²(3x + 5) - 3(3x + 5)

⇒ (x² - 3)(3x + 5)

⇒ $(x - \sqrt{3})(x + \sqrt{3})(3x + 5)$

Hence, a = 4 and 3x³ + 5x² - 9x - 15 = $(x - \sqrt{3})(x + \sqrt{3})(3x + 5)$.

Using remainder theorem, find the value of k if on dividing 2x³ + 3x² - kx + 5 by x - 2, leaves a remainder 7.

Answer

x - 2 = 0 ⇒ x = 2.

Given,

On dividing 2x³ + 3x² - kx + 5 by x - 2, remainder = 7.

∴ 2(2)³ + 3(2)² - k(2) + 5 = 7

⇒ 2(8) + 3(4) - 2k + 5 = 7

⇒ 16 + 12 - 2k + 5 = 7

⇒ 33 - 2k = 7

⇒ 2k = 33 - 7

⇒ 2k = 26

⇒ k = 13.

Hence, k = 13.