Remainder & Factor Theorem — Exercise 8(A)

Using remainder theorem, find the remainder when:

f(x) = 3x² - 5x + 7 is divided by (x - 2).

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

f(x) = 3x² - 5x + 7

Divisor :

⇒ x - 2 = 0

⇒ x = 2

Substituting x = 2 in f(x), we get :

⇒ f(2) = 3(2)² - 5(2) + 7

= 3(4) - 5(2) + 7

= 12 - 10 + 7

= 9.

Hence, remainder = 9.

Using remainder theorem, find the remainder when:

f(x) = 2x³ - 5x² + 3x - 10 is divided by (x - 3).

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

f(x) = 2x³ - 5x² + 3x - 10

Divisor :

⇒ x - 3 = 0

⇒ x = 3

Substituting x = 3 in f(x), we get :

⇒ f(3) = 2(3)³ - 5(3)² + 3(3) - 10

= 2(27) - 5(9) + 3(3) - 10

= 54 - 45 + 9 - 10

= 8.

Hence, remainder = 8.

Using remainder theorem, find the remainder when:

f(x) = 5x³ - 12x² + 17x - 6 is divided by (x - 1).

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

Let f(x) = 5x³ - 12x² + 17x - 6

Divisor :

⇒ x - 1 = 0

⇒ x = 1.

Substituting x = 1 in f(x), we get :

⇒ f(1) = 5(1)³ - 12(1)² + 17(1) - 6

= 5(1) - 12(1) + 17(1) - 6

= 5 - 12 + 17 - 6

= 4.

Hence, remainder = 4.

Using remainder theorem, find the remainder when:

f(x) = x³ - 2x² - 5x + 6 is divided by (x + 2).

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

Let f(x) = x³ - 2x² - 5x + 6

Divisor :

⇒ x + 2 = 0

⇒ x = -2

Substituting x = -2 in f(x), we get :

⇒ f(-2) = (-2)³ - 2(-2)² - 5(-2) + 6

= (-8) - 2(4) - 5(-2) + 6

= -8 - 8 + 10 + 6

= 0.

Hence, remainder = 0.

Using remainder theorem, find the remainder when:

f(x) = 8x³ - 16x² + 14x - 5 is divided by (2x - 1).

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

Let f(x) = 8x³ - 16x² + 14x - 5

Divisor :

⇒ (2x - 1) = 0

⇒ 2x = 1

⇒ x = $\dfrac{1}{2}$

Substituting x = $\dfrac{1}{2}$ in f(x), we get :

$\Rightarrow f\Big(\dfrac{1}{2}\Big) = 8\Big(\dfrac{1}{2}\Big)^3 - 16\Big(\dfrac{1}{2}\Big)^2 + 14\Big(\dfrac{1}{2}\Big) - 5 \\[1em] = 8\Big(\dfrac{1}{8}\Big) - 16\Big(\dfrac{1}{4}\Big) + 14\Big(\dfrac{1}{2}\Big) - 5 \\[1em] = 1 - 4 + 7 - 5 \\[1em] = -1.$

Hence, remainder = -1.

Using remainder theorem, find the remainder when:

f(x) = 9x² - 6x + 2 is divided by (3x - 2).

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

f(x) = 9x² - 6x + 2

Divisor :

⇒ (3x - 2) = 0

⇒ 3x = 2

⇒ x = $\dfrac{2}{3}$

Substituting x = $\dfrac{2}{3}$ in f(x), we get :

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

Hence, remainder = 2.

Using remainder theorem, find the remainder when:

f(x) = 8x² - 2x - 15 is divided by (2x + 3).

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

f(x) = 8x² - 2x - 15

Divisor :

⇒ 2x + 3 = 0

⇒ 2x = -3

⇒ x = -$\dfrac{3}{2}$

Substituting x = -$\dfrac{3}{2}$ in f(x), we get :

$\Rightarrow f\Big(-\dfrac{3}{2}\Big) = 8\Big(-\dfrac{3}{2}\Big)^2 - 2\Big(-\dfrac{3}{2}\Big) - 15 \\[1em] = 8\Big(\dfrac{9}{4}\Big) + 2\Big(\dfrac{3}{2}\Big) - 15 \\[1em] = 18 + 3 - 15 \\[1em] = 6.$

Hence, remainder = 6.

On dividing (ax³ + 9x² + 4x - 10) by (x + 3), we get 5 as remainder. Find the value of a.

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

Let f(x) = ax³ + 9x² + 4x - 10

Given,

Remainder = 5

Divisor :

⇒ x + 3 = 0

⇒ x = -3.

Substituting x = -3 in f(x), will give remainder 5.

⇒ f(-3) = 5

⇒ a(-3)³ + 9(-3)² + 4(-3) - 10 = 5

⇒ a(-27) + 81 - 12 - 10 = 5

⇒ -27a + 81 - 22 = 5

⇒ -27a + 59 = 5

⇒ -27a = 5 - 59

⇒ -27a = -54

⇒ a = $\dfrac{-54}{-27}$

⇒ a = 2.

Hence, the value of a = 2.

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

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

Let f(x) = 2x³ + 3x² - kx + 5

Given,

Remainder = 7

Divisor :

⇒ x - 2 = 0

⇒ x = 2

Substituting x = 2 in f(x), gives remainder 7.

⇒ f(2) = 7

⇒ 2(2)³ + 3(2)² - k(2) + 5 = 7

⇒ 2(8) + 3(4) - 2k + 5 = 7

⇒ 16 + 12 - 2k + 5 = 7

⇒ -2k + 33 = 7

⇒ 2k = 33 - 7

⇒ 2k = 26

⇒ k = $\dfrac{26}{2}$

⇒ k = 13.

Hence, the value of k is 13.

If the polynomials 2x³ + ax² + 3x - 5 and x³ + x² - 2x + a leave the same remainder when divided by (x - 2), find the value of a. Also, find the remainder in each case.

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

Let, p(x) = 2x³ + ax² + 3x - 5 and q(x) = x³ + x² - 2x + a

Divisor :

⇒ x - 2

⇒ x = 2

⇒ p(2) = 2(2)³ + a(2)² + 3(2) - 5

= 2(8) + 4a + 6 - 5

= 16 + 4a + 1

= 4a + 17.

⇒ q(2) = (2)³ + (2)² - 2(2) + a

= 8 + 4 - 4 + a

= 8 + a.

Given,

Polynomials 2x³ + ax² + 3x - 5 and x³ + x² - 2x + a leave the same remainder when divided by (x - 2).

∴ p(2) = q(2)

⇒ 4a + 17 = 8 + a

⇒ 4a - a = 8 - 17

⇒ 3a = -9

⇒ a = $\dfrac{-9}{3}$

⇒ a = -3.

Substituting value of a in p(2) :

⇒ p(2) = 4a + 17

= 4(-3) + 17

= -12 + 17

= 5.

Substituting value of a in q(2) :

⇒ q(2) = 8 + a

= 8 - 3

= 5.

Hence, the value of a = -3 and remainder in each case is 5.

The polynomials f(x) = ax³ + 3x² - 3 and g(x) = 2x³ - 5x + a when divided by (x - 4) leave the same remainder in each case. Find the value of a.

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

Given,

f(x) = ax³ + 3x² - 3

g(x) = 2x³ - 5x + a

Divisor :

⇒ x - 4 = 0

⇒ x = 4

On dividing ax³ + 3x² - 3 by x - 4,

⇒ f(4) = a(4)³ + 3(4)² - 3

= 64a + 48 - 3

= 64a + 45.

On dividing 2x³ - 5x + a by x - 4,

⇒ g(4) = 2(4)³ - 5(4) + a

= 128 - 20 + a

= 108 + a.

Given,

On dividing by (x - 4) polynomials f(x) = ax³ + 3x² - 3 and g(x) = 2x³ - 5x + a leave same remainder.

⇒ f(4) = g(4)

⇒ 64a + 45 = 108 + a

⇒ 64a - a = 108 - 45

⇒ 63a = 63

⇒ a = $\dfrac{63}{63}$

⇒ a = 1.

Hence, the value of a = 1.

Find a if the two polynomials ax³ + 3x² - 9 and 2x³ + 4x + a leave the same remainder when divided by (x + 3).

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

Let p(x) = ax³ + 3x² - 9 and q(x) = 2x³ + 4x + a

Given,

Divisor :

⇒ x + 3 = 0

⇒ x = -3

On dividing ax³ + 3x² - 9 by x + 3, we get :

⇒ p(-3) = a(-3)³ + 3(-3)² - 9

= -27a + 27 - 9

= -27a + 18.

On dividing 2x³ + 4x + a by x + 3, we get :

⇒ q(-3) = 2(-3)³ + 4(-3) + a

= -54 - 12 + a

= -66 + a.

Given,

Polynomials ax³ + 3x² - 9 and 2x³ + 4x + a leave the same remainder when divided by (x + 3).

∴ p(-3) = q(-3)

⇒ -27a + 18 = -66 + a

⇒ -27a - a = -66 - 18

⇒ -28a = -84

⇒ a = $\dfrac{-84}{-28}$

⇒ a = 3.

Hence, the value of a = 3.

If (2x³ + ax² + bx - 2) when divided by (2x - 3) and (x + 3) leaves remainders 7 and -20 respectively, find values of a and b.

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

Let, f(x) = 2x³ + ax² + bx - 2.

Given,

Divisor :

⇒ 2x - 3 = 0

⇒ 2x = 3

⇒ x = $\dfrac{3}{2}$

Given,

On dividing 2x³ + ax² + bx - 2 by 2x - 3, remainder is 7.

$\Rightarrow f\Big(\dfrac{3}{2}\Big) = 7 \\[1em] \Rightarrow 2\Big(\dfrac{3}{2}\Big)^3 + a\Big(\dfrac{3}{2}\Big)^2 + b\Big(\dfrac{3}{2}\Big) - 2 = 7 \\[1em] \Rightarrow 2\Big(\dfrac{27}{8}\Big) + \Big(\dfrac{9a}{4}\Big) + \Big(\dfrac{3b}{2}\Big) - 2 = 7 \\[1em] \Rightarrow \dfrac{27}{4} + \dfrac{9a}{4} + \dfrac{3b \times 2}{4} = 7 + 2 \\[1em] \Rightarrow \Big(\dfrac{27 + 9a + 6b}{4}\Big) = 9 \\[1em] \Rightarrow 27 + 9a + 6b = 9 \times 4 \\[1em] \Rightarrow 27 + 9a + 6b = 36 \\[1em] \Rightarrow 9a + 6b = 36 - 27 \\[1em] \Rightarrow 9a + 6b = 9 \\[1em] \Rightarrow 3(3a + 2b) = 9 \\[1em] \Rightarrow 3a + 2b = \dfrac{9}{3} \\[1em] \Rightarrow 3a + 2b = 3.....(1)$

Divisor :

⇒ x + 3 = 0

⇒ x = -3

On dividing 2x³ + ax² + bx - 2 by x + 3, remainder is -20.

⇒ f(-3) = -20

⇒ 2(-3)³ + a(-3)² + b(-3) - 2 = -20

⇒ 2(-27) + 9a - 3b - 2 = -20

⇒ -54 + 9a - 3b - 2 = -20

⇒ 9a - 3b - 56 = -20

⇒ 9a - 3b = -20 + 56

⇒ 9a - 3b = 36

⇒ 3(3a - b) = 36

⇒ 3a - b = $\dfrac{36}{3}$

⇒ 3a - b = 12

⇒ b = 3a - 12 ....(2)

Substituting value of b from equation (2) in 3a + 2b = 3, we get :

⇒ 3a + 2(3a - 12) = 3

⇒ 3a + 6a - 24 = 3

⇒ 9a = 27

⇒ a = $\dfrac{27}{9}$

⇒ a = 3.

Substituting value of a in equation (2), we get :

⇒ b = 3(3) - 12

⇒ b = 9 - 12

⇒ b = -3.

Hence, the value of a = 3 and b = -3.

Using the Remainder Theorem, find the remainders obtained when x³ + (kx + 8)x + k is divided by x + 1 and x - 2. Hence find k if the sum of the two remainders is 1.

Answer

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

Let f(x) = x³ + (kx + 8)x + k = x³ + kx² + 8x + k

Given,

Divisor :

⇒ x + 1 = 0

⇒ x = -1

On dividing x³ + kx² + 8x + k by x + 1, we get :

⇒ f(-1) = (-1)³ + k(-1)² + 8(-1) + k

= -1 + k - 8 + k

= 2k - 9.

Divisor :

⇒ x - 2 = 0

⇒ x = 2.

On dividing x³ + kx² + 8x + k by x - 2, we get :

⇒ f(2) = (2)³ + k(2)² + 8(2) + k

= 8 + 4k + 16 + k

= 5k + 24

Given,

Sum of two remainders is 1.

⇒ 2k - 9 + 5k + 24 = 1

⇒ 7k + 15 = 1

⇒ 7k = 1 - 15

⇒ k = $-\dfrac{14}{7}$

⇒ k = -2.

Hence, the value of k = -2.