Remainder & Factor Theorem — Analytical & Application Based Questions

Pamela factorized the following polynomial :

2x³ + 3x² - 3x - 2

She found the result as (x + 2)(x - 1)(x - 2). Using remainder and factor theorem, verify whether her result is correct. If incorrect, give the correct result.

Answer

Dividing the polynomial f(x) = 2x³ + 3x² - 3x - 2 by (x + 2), we get :

⇒ x + 2 = 0

⇒ x = -2

f(-2) = 2(-2)³ + 3(-2)² - 3(-2) - 2

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

= -16 + 12 + 6 - 2

= -18 + 18

= 0.

Dividing the polynomial f(x) = 2x³ + 3x² - 3x - 2 by (x - 1), we get :

⇒ x - 1 = 0

⇒ x = 1

f(1) = 2(1)³ + 3(1)² - 3(1) - 2

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

= 2 + 3 - 3 - 2

= 0.

Dividing the polynomial f(x) = 2x³ + 3x² - 3x - 2 by (x - 2), we get :

⇒ x - 2 = 0

⇒ x = 2

f(2) = 2(2)³ + 3(2)² - 3(2) - 2

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

= 16 + 12 - 6 - 2

= 20.

Since, f(2) ≠ 0

∴ x - 2 does not divide the polynomial 2x³ + 3x² - 3x - 2.

Dividing the polynomial f(x) by (x + 2)(x - 1) or by (x² + x - 2), we get :

$\begin{array}{l} \phantom{x^2 + x - 2)}{\quad 2x + 1} \\ x^2 + x - 2\overline{\smash{\big)}\quad 2x^3 + 3x^2 - 3x - 2} \\ \phantom{x^2 + x - 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{-}{+}2x^2 \underset{+}{-}4x} \\ \phantom{{x^2 + x - 2}2x^3 + 20x^3}x^2 + x - 2 \\ \phantom{{x^2 + x - 2)}2x^3 + x^3}\underline{\underset{-}{+}x^2 \underset{-}{+} x \underset{+}{-} 2} \\ \phantom{{x^2 + x - 2)}{x^3-2x^{2}(31)}{x}}\times \end{array}$

∴ 2x³ + 3x² - 3x - 2 = (x² + x - 2)(2x + 1)

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

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

Using remainder and factor theorem, show that (2x + 3) is a factor of the polynomial 2x² + 11x + 12. Hence, factorise it completely. What must be multiplied to the given polynomial so that x² + 3x - 4 is a factor of the resulting polynomial? Also, write the resulting polynomial.

Answer

2x + 3 = 0

⇒ 2x = -3

⇒ x = $-\dfrac{3}{2}$

Substituting value of x in equation 2x² + 11x + 12, we get :

$\Rightarrow 2 \times \Big(-\dfrac{3}{2}\Big)^2 + 11 \times \Big(-\dfrac{3}{2}\Big) + 12 \\[1em] \Rightarrow 2 \times \dfrac{9}{4} - \dfrac{33}{2} + 12 \\[1em] \Rightarrow \dfrac{9}{2} - \dfrac{33}{2} + 12 \\[1em] \Rightarrow \dfrac{9 - 33 + 24}{2} \\[1em] \Rightarrow \dfrac{0}{2} \\[1em] \Rightarrow 0.$

Since, remainder = 0.

∴ 2x + 3 is a factor of the polynomial 2x² + 11x + 12.

Solving polynomial, 2x² + 11x + 12, we get :

⇒ 2x² + 8x + 3x + 12

⇒ 2x(x + 4) + 3(x + 4)

⇒ (2x + 3)(x + 4).

Solving polynomial, x² + 3x - 4, we get :

⇒ x² + 4x - x - 4

⇒ x(x + 4) - 1(x + 4)

⇒ (x - 1)(x + 4).

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

∴ On multiplying polynomial, 2x² + 11x + 12 by (x - 1) it will be divisible by x² + 3x - 4.

⇒ (2x² + 11x + 12)(x - 1)

⇒ 2x³ - 2x² + 11x² - 11x + 12x - 12

⇒ 2x³ + 9x² + x - 12.

Hence, the resulting polynomial = 2x³ + 9x² + x - 12.

Given, 9𝑥² - 4 is a factor of 9𝑥³ - m𝑥² - n𝑥 + 8 :

(i) find the value of m and n using the remainder and factor theorem.

(ii) factorise the given polynomial completely.

Answer

(i) Simplifying 9𝑥² - 4, we get :

⇒ 9𝑥² - 4

⇒ (3𝑥)² - 2²

⇒ (3𝑥 + 2)(3𝑥 - 2).

We know that,

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

Given,

9𝑥² - 4 is a factor of 9𝑥³ - m𝑥² - n𝑥 + 8.

∴ (3𝑥 + 2) and (3𝑥 - 2) are the factors of 9𝑥³ - m𝑥² - n𝑥 + 8.

⇒ 3𝑥 + 2 = 0

⇒ 3𝑥 = -2

⇒ 𝑥 = $-\dfrac{2}{3}$

Substituting x = $-\dfrac{2}{3}$ in 9𝑥³ - mx² - nx + 8, we get remainder = 0.

$\Rightarrow 9 \times \Big(-\dfrac{2}{3}\Big)^3 - m \times \Big(-\dfrac{2}{3}\Big)^2 - n \times \Big(-\dfrac{2}{3}\Big) + 8 = 0 \\[1em] \Rightarrow 9 \times -\dfrac{8}{27} - m \times \dfrac{4}{9} + \dfrac{2n}{3} + 8 = 0 \\[1em] \Rightarrow -\dfrac{8}{3} - \dfrac{4m}{9} + \dfrac{2n}{3} + 8 = 0 \\[1em] \Rightarrow \dfrac{-24 - 4m + 6n + 72}{9} = 0 \\[1em] \Rightarrow \dfrac{6n - 4m + 48}{9} = 0 \\[1em] \Rightarrow 6n - 4m + 48 = 0 \\[1em] \Rightarrow 2(3n - 2m + 24) = 0 \\[1em] \Rightarrow 3n - 2m + 24 = 0 \\[1em] \Rightarrow 3n - 2m = -24 \text{ ..........(1)}$

⇒ 3𝑥 - 2 = 0

⇒ 3𝑥 = 2

⇒ 𝑥 = $\dfrac{2}{3}$

Substituting 𝑥 = $\dfrac{2}{3}$ in 9𝑥³ - m𝑥² - n𝑥 + 8, we get remainder = 0.

$\Rightarrow 9 \times \Big(\dfrac{2}{3}\Big)^3 - m \times \Big(\dfrac{2}{3}\Big)^2 - n \times \Big(\dfrac{2}{3}\Big) + 8 = 0 \\[1em] \Rightarrow 9 \times \dfrac{8}{27} - m \times \dfrac{4}{9} - \dfrac{2n}{3} + 8 = 0 \\[1em] \Rightarrow \dfrac{8}{3} - \dfrac{4m}{9} - \dfrac{2n}{3} + 8 = 0 \\[1em] \Rightarrow \dfrac{24 - 4m - 6n + 72}{9} = 0 \\[1em] \Rightarrow \dfrac{96 - 6n - 4m}{9} = 0 \\[1em] \Rightarrow 96 - 6n - 4m = 0 \\[1em] \Rightarrow 4m + 6n = 96 \\[1em] \Rightarrow 2(2m + 3n) = 96 \\[1em] \Rightarrow 2m + 3n = \dfrac{96}{2} \\[1em] \Rightarrow 2m + 3n = 48 \text{ ..........(2)}$

Adding equation (1) and (2), we get :

⇒ 3n - 2m + 2m + 3n = -24 + 48

⇒ 6n = 24

⇒ n = $\dfrac{24}{6}$

⇒ n = 4.

Substituting value of n in equation (1), we get :

⇒ 3(4) - 2m = -24

⇒ 12 - 2m = -24

⇒ -2m = -24 - 12

⇒ -2m = -36

⇒ m = $\dfrac{-36}{-2}$ = 18.

Hence, m = 18 and n = 4.

(ii) Substituting value of m and n in 9𝑥³ - m𝑥² - n𝑥 + 8, we get :

9𝑥³ - 18𝑥² - 4𝑥 + 8

Dividing 9𝑥³ - 18𝑥² - 4𝑥 + 8 by 9𝑥² - 4, we get :

$\begin{array}{l} \phantom{9x^2 - 4)}{\quad x - 2} \\ 9x^2 - 4\overline{\smash{\big)}\quad 9x^3 - 18x^2 - 4x + 8} \\ \phantom{9x^2 - 4)}\phantom{)}\underline{\underset{-}{+}9x^3 \phantom{- 18x^2} \underset{+}{-}4x} \\ \phantom{{9x^2 - 4}{+9x^3 - }}-18x^2 \phantom{- 4x} + 8 \\ \phantom{{9x^2 - 4)}{+9x^3 - }}\underline{\underset{+}{-}18x^2 \phantom{- 4x} \underset{-}{+} 8} \\ \phantom{{9x^2 - 4)}{x^3-2x^{2}(31)}{x}}\times \end{array}$

∴ 9𝑥³ - 18𝑥² - 4𝑥 + 8 = (9𝑥² - 4)(𝑥 - 2)

= (3𝑥 + 2)(3𝑥 - 2)(𝑥 - 2).

Hence, factors are (3𝑥 + 2), (3𝑥 - 2) and (𝑥 - 2).

If a polynomial x³ + 2x² – ax + b leaves a remainder -6 when divided by x + 1 and the same polynomial has x - 2 as a factor, then find the values of a and b.

Answer

Given,

x³ + 2x² – ax + b leaves a remainder -6 when divided by x + 1.

⇒ x + 1 = 0

⇒ x = -1.

Let f(x) = x³ + 2x² – ax + b, then f(-1) = -6.

⇒ f(-1) = -6

⇒ (-1)³ + 2(-1)² - a(-1) + b = -6

⇒ -1 + 2(1) + a + b = -6

⇒ -1 + 2 + a + b = -6

⇒ a + b + 1 = -6

⇒ a + b = -6 - 1

⇒ a + b = -7 ........(1)

Given ,

x - 2 is a factor of f(x).

⇒ x - 2 = 0

⇒ x = 2.

f(2) = 0

⇒ 2³ + 2(2)² – 2a + b = 0

⇒ 8 + 2(4) - 2a + b = 0

⇒ 8 + 8 - 2a + b = 0

⇒ 16 - 2a + b = 0

⇒ 2a - b = 16 ......(2)

Adding equation (1) and (2), we get :

⇒ a + b + 2a - b = -7 + 16

⇒ 3a = 9

⇒ a = $\dfrac{9}{3}$ = 3.

Substituting value of a = 3 in equation (1), we get :

⇒ 3 + b = -7

⇒ b = -7 - 3 = -10.

Hence, a = 3 and b = -10.