Remainder & Factor Theorem — Exercise 8(B)

Using factor theorem, show that:

(x - 3) is a factor of (x³ + x² - 17x + 15).

Answer

Let f(x) = (x³ + x² - 17x + 15).

Given,

Divisor :

⇒ x - 3 = 0

⇒ x = 3

By factor theorem,

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

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

⇒ f(3) = (3)³ + (3)² - 17(3) + 15

= 27 + 9 - 51 + 15

= 51 - 51

= 0.

Since, f(3) = 0, thus (x - 3) is a factor of f(x).

Hence, proved that (x - 3) is factor of x³ + x² - 17x + 15.

Using factor theorem, show that:

(x + 1) is a factor of (x³ + 4x² + 5x + 2).

Answer

Let f(x) = (x³ + 4x² + 5x + 2)

Given,

Divisor:

⇒ x + 1 = 0

⇒ x = -1

By factor theorem,

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

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

⇒ f(-1) = (-1)³ + 4(-1)² + 5(-1) + 2

= -1 + 4(1) - 5 + 2

= -1 + 4 - 5 + 2

= 6 - 6

= 0.

Since, f(-1) = 0, thus (x + 1) is a factor of f(x).

Hence, proved that (x + 1) is factor of x³ + 4x² + 5x + 2.

Using factor theorem, show that:

(3x - 2) is a factor of (3x³ + x² - 20x + 12).

Answer

Let f(x) = (3x³ + x² - 20x + 12)

Given,

Divisor:

⇒ 3x - 2 = 0

⇒ 3x = 2

⇒ x = $\dfrac{2}{3}$

By factor theorem,

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

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

$\Rightarrow f\Big(\dfrac{2}{3}\Big) = 3\Big(\dfrac{2}{3}\Big)^3 + \Big(\dfrac{2}{3}\Big)^2 - 20\Big(\dfrac{2}{3}\Big) + 12 \\[1em] = \Big(\dfrac{8}{9}\Big) + \Big(\dfrac{4}{9}\Big) - \Big(\dfrac{40}{3}\Big) + 12 \\[1em] = \Big(\dfrac{8 + 4 - 120 + 108}{9}\Big) \\[1em] = \Big(\dfrac{120 - 120}{9}\Big) \\[1em] = 0.$

Since, $f\Big(\dfrac{2}{3}\Big)$ = 0, thus (3x - 2) is a factor of f(x).

Hence, proved that (3x - 2) is factor of 3x³ + x² - 20x + 12.

Using factor theorem, show that:

(3 - 2x) is a factor of (2x³ - 9x² + x + 12).

Answer

Let f(x) = (2x³ - 9x² + x + 12)

Given,

Divisor:

⇒ 3 - 2x = 0

⇒ 2x = 3

⇒ x = $\dfrac{3}{2}$

By factor theorem,

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

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

$\Rightarrow f\Big(\dfrac{3}{2}\Big) = 2\Big(\dfrac{3}{2}\Big)^3 - 9\Big(\dfrac{3}{2}\Big)^2 + \Big(\dfrac{3}{2}\Big) + 12 \\[1em] = 2\Big(\dfrac{27}{8}\Big) - 9\Big(\dfrac{9}{4}\Big) + \Big(\dfrac{3}{2}\Big) + 12 \\[1em] = \dfrac{27}{4} - \dfrac{81}{4} + \dfrac{3}{2} + 12 \\[1em] = \Big(\dfrac{27 - 81 + 6 + 48}{4}\Big) \\[1em] = \dfrac{81 - 81}{27} \\[1em] = 0.$

Since, f$\Big(\dfrac{3}{2}\Big)$ = 0, thus (3 - 2x) is a factor of f(x).

Hence, proved that (3 - 2x) is factor of 2x³ - 9x² + x + 12.

Use factor theorem to show that (x + 2) and (2x - 3) are factors of (2x² + x - 6).

Answer

Let f(x) = 2x² + x - 6

Given,

Factors : (x + 2) and (2x - 3)

⇒ x + 2 = 0 and 2x - 3 = 0

⇒ x = -2 and 2x = 3

⇒ x = -2 and x = $\dfrac{3}{2}$

By factor theorem,

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

Thus, (x + 2) and (2x - 3) are factors of f(x), if f(-2) = 0 and f$\Big(\dfrac{3}{2}\Big)$ = 0.

On dividing 2x² + x - 6 by (x + 2), we get :

⇒ f(-2) = 2(-2)² + (-2) - 6

= 2(4) - 2 - 6

= 8 - 2 - 6

= 8 - 8

= 0.

On dividing 2x² + x - 6 by (2x - 3), we get :

$\Rightarrow f\Big(\dfrac{3}{2}\Big) = 2\Big(\dfrac{3}{2}\Big)^2 + \Big(\dfrac{3}{2}\Big) - 6 \\[1em] = 2\Big(\dfrac{9}{4}\Big) + \Big(\dfrac{3}{2}\Big) - 6 \\[1em] = \Big(\dfrac{9}{2}\Big) + \Big(\dfrac{3}{2}\Big) - 6 \\[1em] = \Big(\dfrac{9 + 3}{2}\Big) - 6 \\[1em] = \Big(\dfrac{12}{2}\Big) - 6 \\[1em] = 6 - 6 \\[1em] = 0.$

Since, f(-2) = $f\Big(\dfrac{3}{2}\Big)$ = 0.

Hence, proved that x + 2 and 2x - 3 are factors of 2x² + x - 6.

Find the value of a so that (x + 6) is a factor of the polynomial (x³ + 5x² - 4x + a).

Answer

Let f(x) = x³ + 5x² - 4x + a.

Given,

Factor: x + 6

Thus, on dividing x³ + 5x² - 4x + a by x + 6, remainder will be zero.

⇒ f(-6) = 0

⇒ (-6)³ + 5(-6)² - 4(-6) + a = 0

⇒ -216 + 180 + 24 + a = 0

⇒ -12 + a = 0

⇒ a = 12.

Hence, the value of a = 12.

For what value of a is the polynomial (2x³ + ax² + 11x + a + 3) exactly divisible by (2x - 1)?

Answer

Let f(x) = 2x³ + ax² + 11x + a + 3.

Given,

Factor : 2x - 1

⇒ 2x - 1 = 0

⇒ 2x = 1

⇒ x = $\dfrac{1}{2}$

Thus, on dividing 2x³ + ax² + 11x + a + 3 by 2x - 1, remainder = 0.

$\therefore f\Big(\dfrac{1}{2}\Big) = 0 \\[1em] \Rightarrow 2\Big(\dfrac{1}{2}\Big)^3 + a\Big(\dfrac{1}{2}\Big)^2 + 11\Big(\dfrac{1}{2}\Big) + a + 3 = 0 \\[1em] \Rightarrow 2\Big(\dfrac{1}{8}\Big) + \Big(\dfrac{a}{4}\Big) + \Big(\dfrac{11}{2}\Big) + a + 3 = 0 \\[1em] \Rightarrow \Big(\dfrac{1}{4}\Big) + \Big(\dfrac{a}{4}\Big) + \Big(\dfrac{11}{2}\Big) + a + 3 = 0 \\[1em] \Rightarrow \Big(\dfrac{1 + a + 22 + 4a + 12}{4}\Big) = 0 \\[1em] \Rightarrow 1 + a + 22 + 4a + 12 = 0 \\[1em] \Rightarrow 5a + 35 = 0 \\[1em] \Rightarrow 5a = -35 \\[1em] \Rightarrow a = \dfrac{-35}{5} = -7.$

Hence, the value of a = -7.

What must be subtracted from the polynomial x³ + x² - 2x + 1, so that the result is exactly divisible by (x - 3)?

Answer

Polynomial : x³ + x² - 2x + 1

Division by x - 3

⇒ x - 3 = 0

⇒ x = 3.

Let k be subtracted from the polynomial, so resulting polynomial is x³ + x² - 2x + 1 - k.

Resulting polynomial should be exactly divisible by x - 3,

Thus, substituting x = 3, in polynomial x³ + x² - 2x + 1 - k, remainder = 0.

⇒ 3³ + 3² - 2(3) + 1 - k = 0

⇒ 27 + 9 - 6 + 1 - k = 0

⇒ 31 - k = 0

⇒ k = 31.

Hence, k = 31.

What must be subtracted from 16x³ - 8x² + 4x + 7 so that the resulting expression has (2x + 1) as a factor?

Answer

Let the number to be subtracted from 16x³ - 8x² + 4x + 7 be a.

Resulting polynomial [f(x)] = 16x³ - 8x² + 4x + 7 - a

Given,

Factor: 2x + 1

⇒ 2x + 1 = 0

⇒ 2x = -1

⇒ x = $-\dfrac{1}{2}$.

Since, 2x + 1 is a factor.

Thus, on dividing 16x³ - 8x² + 4x + 7 - a by 2x + 1, remainder = 0.

$\therefore f\Big(\dfrac{-1}{2}\Big) = 0 \\[1em] \Rightarrow 16\Big(\dfrac{-1}{2}\Big)^3 - 8\Big(\dfrac{-1}{2}\Big)^2 + 4\Big(\dfrac{-1}{2}\Big) + 7 - a = 0 \\[1em] \Rightarrow 16\Big(\dfrac{-1}{8}\Big) - 8\Big(\dfrac{1}{4}\Big) - 2 + 7 - a = 0 \\[1em] \Rightarrow -2 - 2 - 2 + 7 - a = 0 \\[1em] \Rightarrow -6 + 7 - a = 0 \\[1em] \Rightarrow a = 1.$

Hence, the required number to be subtracted from the polynomial = 1.

Using factor theorem, show that (x - 3) is a factor of (x³ - 7x² + 15x - 9). Hence, factorize the given expression completely.

Answer

Let f(x) = x³ - 7x² + 15x - 9.

We know that,

(x - 3) will be the factor of f(x), if f(3) will be equal to 0.

⇒ f(3) = (3)³ - 7(3)² + 15(3) - 9.

= 27 - 63 + 45 - 9

= 72 - 72

= 0.

Since, f(3) = 0, thus (x - 3) is a factor of f(x).

Now, dividing f(x) by x - 3,

$\begin{array}{l} \phantom{x - 3)}{x^2 - 4x + 3} \\ x - 3\overline{\smash{\big)}x^3 - 7x^2 + 15x - 9} \\ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-}3x^2} \\ \phantom{{x - 2}x^3-}-4x^2 + 15x \\ \phantom{{x - 2}x^3ok\space}\underline{\underset{+}{-}4x^2\underset{-}{+} 12x} \\ \phantom{{x - 2uo[ki]}x^3okklk\space}{3x - 9} \\ \phantom{{x - 2}x^3o;lmkb\k\space}\underline{\underset{-}{+}3x\underset{+}{-} 9} \\ \phantom{{x - 2}{x^,jo\3-2x^2\space}{-9x}}\times \end{array}$

∴ x³ - 7x² + 15x - 9 = (x - 3)(x² - 4x + 3)

= (x - 3)(x² - 3x - x + 3)

= (x - 3)[x(x - 3) - 1(x - 3)]

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

= (x - 3)²(x - 1).

Hence, x³ - 7x² + 15x - 9 = (x - 3)²(x - 1).

Using factor theorem, show that (x - 4) is a factor of (2x³ + x² - 26x - 40) and hence factorize (2x³ + x² - 26x - 40).

Answer

Let f(x) = 2x³ + x² - 26x - 40

Substituting x = 4 in f(x), we get :

f(4) = 2(4)³ + (4)² - 26(4) - 40

= 2(64) + 16 - 104 - 40

= 128 + 16 - 104 - 40

= 144 - 144

= 0.

Since f(4) = 0, (x − 4) is a factor of 2x³ + x² - 26x - 40

Now, dividing f(x) by x - 4,

$\begin{array}{l} \phantom{x - 3)}{2x^2 + 9x + 10} \\ x - 4\overline{\smash{\big)}2x^3 + x^2 - 26x - 40} \\ \phantom{x - 2}\underline{\underset{-}{+}2x^3 \underset{+}{-}8x^2} \\ \phantom{{x - 2}x^3-}9x^2 - 26x \\ \phantom{{x -l2}x^3\space}\underline{\underset{-}{+}9x^2\underset{+}{-} 36x} \\ \phantom{{x - 2uo[ki]}x^3okk\space}{10x - 40} \\ \phantom{{x - 2}x^3o;lmkk\space}\underline{\underset{-}{+}10x\underset{+}{-} 40} \\ \phantom{{x - 2}{x^,j-k2x^2\space}{-9x}}\times \end{array}$

∴ 2x³ + x² - 26x - 40 = (x - 4)(2x² + 9x + 10)

= (x - 4)(2x² + 4x + 5x + 10)

= (x - 4)[2x(x + 2) + 5(x + 2)]

= (x - 4)(2x + 5)(x + 2)

Hence, 2x³ + x² - 26x - 40 = (x - 4)(2x + 5)(x + 2).

Show that (x - 3) is a factor of (2x³ - 3x² - 11x + 6) and hence factorize (2x³ - 3x² - 11x + 6).

Answer

Let f(x) = 2x³ - 3x² - 11x + 6

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

f(3) = 2(3)³ - 3(3)² - 11(3) + 6

= 54 - 27 - 33 + 6

= 60 - 60

= 0.

Since f(3) = 0, thus (x − 3) is a factor of 2x³ - 3x² - 11x + 6.

Now, dividing f(x) by x - 3,

$\begin{array}{l} \phantom{x - 3)}{2x^2 + 3x - 2} \\ x - 3\overline{\smash{\big)}2x^3 - 3x^2 - 11x + 6} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}6x^2} \\ \phantom{{x - 2}x^,,,3-}3x^2 - 11x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}3x^2 \underset{+}{-} 9x} \\ \phantom{{x - 2euo[ki]}x^3okk\space}{-2x + 6} \\ \phantom{{x - 2}x^3o;llk]lmk,\space}\underline{\underset{+}{-}2x\underset{-}{+} 6} \\ \phantom{{x - 2}{x^,jo\ -k2x^2\space}{-9x}}\times \end{array}$

∴ 2x³ - 3x² - 11x + 6 = (x - 3)(2x² + 3x - 2)

= (x - 3)(2x² + 4x - x - 2)

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

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

Hence, 2x³ - 3x² - 11x + 6 = (x - 3)(2x - 1)(x + 2).

Show that (3x + 2) is a factor of (6x³ + 13x² - 4) and hence factorize (6x³ + 13x² - 4).

Answer

Let, f(x) = 6x³ + 13x² - 4.

Factor :

⇒ 3x + 2 = 0

⇒ 3x = -2

⇒ x = $-\dfrac{2}{3}$.

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

$f\Big(-\dfrac{2}{3}\Big) = 6\Big(-\dfrac{2}{3}\Big)^3 + 13\Big(-\dfrac{2}{3}\Big)^2 - 4 \\[1em] = 6\Big(-\dfrac{8}{27}\Big) + 13\Big(\dfrac{4}{9}\Big) - 4 \\[1em] = \Big(-\dfrac{48}{27}\Big) + \Big(\dfrac{52}{9}\Big) - 4 \\[1em] = \dfrac{-48 + 156 - 108}{27} \\[1em] = \dfrac{156 - 156}{27} \\[1em] = 0.$

Since, $f\Big(-\dfrac{2}{3}\Big)$ = 0, thus (3x + 2) is a factor of 6x³ + 13x² - 4.

Now, dividing f(x) by (3x + 2),

$\begin{array}{l} \phantom{x -;]/3)}{2x^2 + 3x - 2} \\ 3x + 2\overline{\smash{\big)}6x^3 + 13x^2 - 4} \\ \phantom{x - 2l}\underline{\underset{-}{}6x^3 \underset{-}{+}4x^2} \\ \phantom{{x - 2}ex^,,,3-}9x^2 \\ \phantom{{x e[[-l2}fx^3]\space}\underline{\underset{-}{+}9x^2 \underset{-}{+} 6x} \\ \phantom{{x - 2]euo[ki]}x^3ok\space}{-6x - 4} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{+}{-}6x\underset{+}{-} 4} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

6x³ + 13x² - 4 = (3x + 2)(2x² + 3x - 2)

= (3x + 2)(2x² + 4x - x - 2)

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

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

Hence, 6x³ + 13x² - 4 = (3x + 2)(2x - 1)(x + 2).

If 2x³ - 3x² - 3x + 2 = (2x - 1)(x² + ax + b)

(i) using Remainder and Factor theorem, find the value of ‘a’ and ‘b’.

(ii) hence, factorise the polynomial 2x³ - 3x² - 3x + 2 completely.

Answer

(i) Given,

2x³ - 3x² - 3x + 2 = (2x - 1)(x² + ax + b)

Therefore, (2x - 1) is factor of 2x³ - 3x² - 3x + 2.

Factorizing,

$\begin{array}{l} \phantom{x - 3x)}{\quad x^2 - x - 2} \\ 2x - 1\overline{\smash{\big)}\quad 2x^3 - 3x^2 - 3x + 2 } \\ \phantom{x^2 + 4}\phantom(\underline{\underset{-}{+}2x^3 \underset{+}{-}x^2} \\ \phantom{x^2 + 3x - 7)} - 2x^2 - 3x \\ \phantom{x^2 + 3x - 5))}\underline{\underset{+}{-}2x^2 \underset{-}{+}x} \\ \phantom{x^2 + 3x - 5) + 24)}-4x + 2 \\ \phantom{{x^2 + 3x - 5) + 24 +}}\underline{\underset{+}{-}4x \underset{-}{+}2} \\ \phantom{{x^2 + 3x - 54)} + 2x + 7} \times\ \end{array}$

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

Comparing, x² + ax + b with x² - x - 2, we get:

a = -1 and b = -2.

Hence, a = -1 and b = -2.

(ii) From part (i),

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

= (2x - 1)(x² - 2x + x - 2)

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

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

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

While factorizing a given polynomial, using remainder and factor theorem, a student finds that (2x + 1) is a factor of 2x³ + 7x² + 2x – 3.

(i) Is the student’s solution correct stating that (2x + 1) is a factor of the given polynomial ?

(ii) Give a valid reason for your answer.

Also factorize the given polynomial completely.

Answer

Given,

⇒ 2x + 1 = 0

⇒ 2x = -1

⇒ x = $-\dfrac{1}{2}$

Substituting x = $-\dfrac{1}{2}$ in 2x³ + 7x² + 2x – 3, we get:

$\Rightarrow 2 \times \Big(-\dfrac{1}{2}\Big)^3 + 7 \times \Big(-\dfrac{1}{2}\Big)^2 + 2 \times \Big(-\dfrac{1}{2}\Big) - 3 \\[1em] \Rightarrow 2 \times -\dfrac{1}{8} + 7 \times \dfrac{1}{4} + (-1) - 3 \\[1em] \Rightarrow -\dfrac{1}{4} + \dfrac{7}{4} - 4 \\[1em] \Rightarrow \dfrac{-1 + 7}{4} - 4 \\[1em] \Rightarrow \dfrac{6}{4} - 4 \\[1em] \Rightarrow \dfrac{6 - 16}{4} \\[1em] \Rightarrow \dfrac{-10}{4} \\[1em] \Rightarrow \dfrac{-5}{2}.$

Since, remainder is not equal to zero.

Hence, (2x + 1) is not a factor of the given polynomial.

Substituting x = $\dfrac{1}{2}$ in 2x³ + 7x² + 2x – 3, we get:

$\Rightarrow 2 \times \Big(\dfrac{1}{2}\Big)^3 + 7 \times \Big(\dfrac{1}{2}\Big)^2 + 2 \times \Big(\dfrac{1}{2}\Big) - 3 \\[1em] \Rightarrow 2 \times \dfrac{1}{8} + 7 \times \dfrac{1}{4} + 1 - 3 \\[1em] \Rightarrow \dfrac{1}{4} + \dfrac{7}{4} - 2 \\[1em] \Rightarrow \dfrac{1 + 7}{4} - 2 \\[1em] \Rightarrow \dfrac{8}{4} - 2 \\[1em] \Rightarrow 2 - 2 \\[1em] \Rightarrow 0.$

Since, remainder is equal to zero. Thus, x - $\dfrac{1}{2}$ is the factor of the polynomial.

$\Rightarrow x - \dfrac{1}{2} = 0 \\[1em] \Rightarrow x = \dfrac{1}{2} \\[1em] \Rightarrow 2x = 1 \\[1em] \Rightarrow 2x - 1.$

2x - 1 is factor of polynomial.

Dividing 2x³ + 7x² + 2x – 3 by 2x - 1, we get:

$\begin{array}{l} \phantom{x - 3x)}{\quad x^2 + 4x + 3} \\ 2x - 1\overline{\smash{\big)}\quad 2x^3 + 7x^2 + 2x – 3} \\ \phantom{x^2 +o 4}\phantom(\underline{\underset{-}{+}2x^3 \underset{+}{-}x^2} \\ \phantom{x^2 + 3x - 7)[][]} 8x^2 + 2x \\ \phantom{x^2 + 3x - 5))}\underline{\underset{-}{+}8x^2 \underset{+}{-}4x} \\ \phantom{x^2 + 3x - 5) + (24)}6x - 3 \\ \phantom{{x^2 + 3x - 5) + 24}}\underline{\underset{-}{+}6x \underset{+}{-}3} \\ \phantom{{x^2 + 3x - 54)} + 2x ,k}\times\ \end{array}$

2x³ + 7x² + 2x – 3 by 2x - 1 = (2x - 1)(x² + 4x + 3)

= (2x - 1)[x² + 3x + x + 3]

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

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

Hence, 2x³ + 7x² + 2x – 3 = (2x - 1)(x + 1)(x + 3).

Using the factor theorem, show that (x - 2) is a factor of x³ + x² - 4x - 4. Hence factorize the polynomial completely.

Answer

Let, f(x) = x³ + x² - 4x - 4.

Factor :

⇒ x - 2 = 0

⇒ x = 2.

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

⇒ f(2) = (2)³ + (2)² - 4(2) - 4

= 8 + 4 - 8 - 4

= 12 - 12

= 0.

Since, f(2) = 0, thus (x - 2) is a factor of (x³ + x² - 4x - 4).

Now, dividing f(x) by (x - 2), we get :

$\begin{array}{l} \phantom{x -13}{x^2 + 3x + 2} \\ x - 2\overline{\smash{\big)}x^3 + x^2 - 4x - 4} \\ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-}2x^2} \\ \phantom{{x - 2}x^,.3-}3x^2 - 4x \\ \phantom{{x -l2}fx^l.\space}\underline{\underset{-}{+}3x^2 \underset{+}{-}6x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{2x - 4} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}2x\underset{+}{-}4} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

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

= (x - 2)(x² + x + 2x + 2)

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

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

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

If (x - 2) is a factor of 2x³ - x² - px - 2,

(i) find the value of p

(ii) with the value of p, factorize the above expression completely.

Answer

(i) Let f(x) = 2x³ - x² - px - 2

Since, (x − 2) is the factor, f(2) = 0.

⇒ 2(2)³ - (2)² - p(2) - 2 = 0

⇒ 16 - 4 - 2p - 2 = 0

⇒ 10 - 2p = 0

⇒ 2p = 10

⇒ p = $\dfrac{10}{2}$

⇒ p = 5.

Hence, the value of p = 5.

(ii) f(x) = 2x³ - x² - 5x - 2

Now, dividing f(x) by (x - 2), we get :

$\begin{array}{l} \phantom{x -1 ]3)}{2x^2 + 3x + 1} \\ x - 2\overline{\smash{\big)}2x^3 - x^2 - 5x - 2} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}4x^2} \\ \phantom{{x - 2}x^,.3-}3x^2 - 5x \\ \phantom{{x -l2}fx^l.\space}\underline{\underset{-}{+}3x^2 \underset{+}{-}6x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{x - 2} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}x\underset{+}{-}2} \\ \phantom{{x - 2}{x^,jo-k2\space}{-9x}}\times \end{array}$

2x³ - x² - 5x - 2 = (x - 2)(2x² + 3x + 1)

= (x - 2)(2x² + 2x + x + 1)

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

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

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

Find the value of a, if (x - a) is a factor of the polynomial 3x³ + x² - ax - 81.

Answer

Let, f(x) = 3x³ + x² - ax - 81.

Factor :

⇒ x - a = 0

⇒ x = a.

Since (x − a) is the factor, thus f(a) = 0.

⇒ 3(a)³ + a² - a(a) - 81 = 0

⇒ 3a³ + a² - a² - 81 = 0

⇒ 3a³ - 81 = 0

⇒ 3a³ = 81

⇒ a³ = $\dfrac{81}{3}$

⇒ a³ = 27

⇒ a = $\sqrt[3]{27}$

⇒ a = 3.

Hence, the value of a = 3.

Find the values of a and b, if (x - 1) and (x + 2) are both factors of (x³ + ax² + bx - 6).

Answer

Let f(x) = x³ + ax² + bx - 6

Since (x − 1) and (x + 2) are factors, by the factor theorem, f(1) = 0 and f(−2) = 0.

⇒ f(1) = 0

⇒ (1)³ + a(1)² + b(1) - 6 = 0

⇒ 1 + a + b - 6 = 0

⇒ a + b - 5 = 0

⇒ a + b = 5 ....(1)

⇒ f(-2) = 0

⇒ (-2)³ + a(-2)² + b(-2) - 6 = 0

⇒ -8 + 4a - 2b - 6 = 0

⇒ 4a - 2b = 14

⇒ 2(2a - b) = 14

⇒ 2a - b = $\dfrac{14}{2}$

⇒ 2a - b = 7 ....(2)

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

⇒ a + b + 2a - b = 5 + 7

⇒ 3a = 12

⇒ a = $\dfrac{12}{3}$

⇒ a = 4.

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

⇒ 4 + b = 5

⇒ b = 5 - 4

⇒ b = 1.

Hence, the value of a = 4 and b = 1.

If (x + 2) and (x + 3) are factors of x³ + ax + b, find the values of a and b.

Answer

Let f(x) = x³ + ax + b

Since (x + 2) and (x + 3) are factors, by the factor theorem, f(−2) = 0 and f(−3) = 0.

⇒ f(-2) = 0

⇒ (-2)³ + a(-2) + b = 0

⇒ -8 - 2a + b = 0

⇒ -2a + b = 8 ....(1)

⇒ f(-3) = 0

⇒ (-3)³ + a(-3) + b = 0

⇒ -27 - 3a + b = 0

⇒ -3a + b = 27 ....(2)

Subtract equation (2) from equation (1), we get:

⇒ -2a + b - (-3a + b) = 8 - 27

⇒ -2a + 3a = -19

⇒ a = -19

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

⇒ -2(-19) + b = 8

⇒ 38 + b = 8

⇒ b = 8 - 38

⇒ b = -30

Hence, the value of a = -19 and b = -30.

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

Let f(x) = x³ + ax² + bx + 6

By factor theorem,

If, (x - 2) is a factor of f(x), then f(2) = 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

⇒ 2a + b = -7 ....(1)

Given,

On dividing f(x) by (x − 3), the remainder is 3.

By remainder theorem,

∴ f(3) = 3

⇒ (3)³ + a(3)² + b(3) + 6 = 3

⇒ 27 + 9a + 3b + 6 = 3

⇒ 9a + 3b + 33 = 3

⇒ 9a + 3b = 3 - 33

⇒ 9a + 3b = -30

⇒ 3(3a + b) = -30

⇒ 3a + b = $-\dfrac{30}{3}$

⇒ 3a + b = -10 ....(2)

Subtracting equation (1) from equation (2),

⇒ 3a + b - (2a + b) = -10 -(-7)

⇒ 3a + b - 2a - b = -10 + 7

⇒ a = -3.

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

⇒ 2(-3) + b = -7

⇒ -6 + b = -7

⇒ b = -7 + 6

⇒ b = -1.

Hence, the value of a = -3 and b = -1.

Using factor theorem, factorize the following:

x³ + 7x² + 7x - 15

Answer

Let, f(x) = x³ + 7x² + 7x - 15.

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

f(1) = (1)³ + 7(1)² + 7(1) - 15

= 1 + 7 + 7 - 15

= 0.

Since, f(1) = 0, thus (x - 1) is a factor of f(x).

Dividing, f(x) by (x - 1), we get :

$\begin{array}{l} \phantom{x - ]3)}{x^2 + 8x + 15} \\ x - 1\overline{\smash{\big)}x^3 + 7x^2 + 7x - 15} \\ \phantom{x - 2l}\underline{\underset{-}{}x^3 \underset{+}{-}x^2} \\ \phantom{{x - 2}x^,,,3-}8x^2 + 7x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}8x^2 \underset{+}{-} 8x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{15x - 15} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}15x\underset{+}{-} 15} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

∴ x³ + 7x² + 7x - 15 = (x - 1)(x² + 8x + 15)

= (x - 1)(x² + 3x + 5x + 15)

= (x - 1)[x(x + 3) + 5(x + 3)]

= (x - 1)(x + 5)(x + 3).

Hence, x³ + 7x² + 7x - 15 = (x - 1)(x + 5)(x + 3).

Using factor theorem, factorize the following:

6x³ - 7x² - 11x + 12

Answer

Let, f(x) = 6x³ - 7x² - 11x + 12.

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

f(1) = 6(1)³ - 7(1)² - 11(1) + 12

= 6 - 7 - 11 + 12

= 0

Since, f(1) = 0, (x - 1) is a factor of f(x).

Dividing f(x) by (x - 1), we get :

$\begin{array}{l} \phantom{x - ]3)}{6x^2 - x - 12} \\ x - 1\overline{\smash{\big)}6x^3 - 7x^2 - 11x + 12} \\ \phantom{x - 2l}\underline{\underset{-}{}6x^3 \underset{+}{-}6x^2} \\ \phantom{{x - 2}x^,,,3-}-x^2 - 11x \\ \phantom{{x -k.l2}fx^3]\space}\underline{\underset{+}{-}x^2 \underset{-}{+}x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{-12x + 12} \\ \phantom{{x - 2}x^3o;llk]lmk,\space}\underline{\underset{+}{-}12x\underset{-}{+}12} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

∴ 6x³ − 7x² − 11x + 12 = (x − 1)(6x² − x − 12)

= (x − 1)(6x² − 9x + 8x − 12)

= (x − 1)[3x(2x − 3) + 4(2x − 3)]

= (x − 1)(2x − 3)(3x + 4)

Hence, 6x³ − 7x² − 11x + 12 = (x − 1)(2x − 3)(3x + 4).

Using factor theorem, factorize the following:

2x³ + 3x² − 9x − 10

Answer

Let, f(x) = 2x³ + 3x² − 9x − 10.

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

f(2) = 2(2)³ + 3(2)² − 9(2) − 10

= 16 + 12 − 18 − 10

= 0.

Since, f(2) = 0, (x − 2) is a factor of f(x).

Dividing f(x) by (x − 2), we get :

$\begin{array}{l} \phantom{x - ]k3)}{2x^2 + 7x + 5} \\ x - 2\overline{\smash{\big)}2x^3 + 3x^2 - 9x - 10} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}4x^2} \\ \phantom{{x - 2}x^,,,3-}7x^2 - 9x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}7x^2 \underset{+}{-}14x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{5x - 10} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}5x\underset{+}{-}10} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

∴ 2x³ + 3x² − 9x − 10 = (x − 2)(2x² + 7x + 5)

= (x − 2)(2x² + 5x + 2x + 5)

= (x − 2)[x(2x + 5) + 1(2x + 5)]

= (x − 2)(2x + 5)(x + 1)

Hence, 2x³ + 3x² − 9x − 10 = (x − 2)(2x + 5)(x + 1).

Using factor theorem, factorize the following:

2x³ + 19x² + 38x + 21

Answer

Let, f(x) = 2x³ + 19x² + 38x + 21.

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

f(-1) = 2(-1)³ + 19(-1)² + 38(-1) + 21

= 2(-1) + 19(1) - 38 + 21

= -2 + 19 - 38 + 21

= 0.

Since, f(−1) = 0, (x + 1) is a factor of f(x).

Dividing f(x) by (x + 1), we get :

$\begin{array}{l} \phantom{x - ]3)}{2x^2 + 17x + 21} \\ x + 1\overline{\smash{\big)}2x^3 + 19x^2 + 38x + 21} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{-}{+}2x^2} \\ \phantom{{x - 2}x^,,,3-}17x^2 + 38x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}17x^2 \underset{-}{+}17x} \\ \phantom{{x - ll2]euo[ki]}x^3okk\space}{21x + 21} \\ \phantom{{x - 2}x^3o;lklk]lmk\space}\underline{\underset{-}{+}21x\underset{-}{+} 21} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

∴ 2x³ + 19x² + 38x + 21 = (x + 1)(2x² + 17x + 21)

= (x + 1)(2x² + 14x + 3x + 21)

= (x + 1)[2x(x + 7) + 3(x + 7)]

= (x + 1)(x + 7)(2x + 3)

Hence, 2x³ + 19x² + 38x + 21 = (x + 1)(x + 7)(2x + 3).

Using factor theorem, factorize the following:

3x³ + 2x² - 19x + 6

Answer

Let, f(x) = 3x³ + 2x² - 19x + 6.

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

f(2) = 3(2)³ + 2(2)² - 19(2) + 6

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

= 24 + 8 - 38 + 6

= 0.

Since, f(2) = 0, thus (x - 2) is a factor of f(x).

Dividing f(x) by (x - 2), we get :

$\begin{array}{l} \phantom{x -.]3)}{3x^2 + 8x - 3} \\ x - 2\overline{\smash{\big)}3x^3 + 2x^2 - 19x + 6} \\ \phantom{x - l}\underline{\underset{-}{}3x^3 \underset{+}{-}6x^2} \\ \phantom{{x - 2}x^,,,3-}8x^2 - 19x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}8x^2 \underset{+}{-}16x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{-3x + 6} \\ \phantom{{x - 2}x,'^3o;llk]lmk\space}\underline{\underset{+}{-}3x\underset{-}{+}6} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

∴ 3x³ + 2x² - 19x + 6 = (x - 2)(3x² + 8x - 3)

= (x - 2)(3x² + 9x - x - 3)

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

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

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

Using factor theorem, factorize the following:

2x³ + x² - 13x + 6

Answer

Let, f(x) = 2x³ + x² - 13x + 6.

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

f(2) = 2(2)³ + (2)² - 13(2) + 6

= 2(8) + 4 - 26 + 6

= 16 + 4 - 26 + 6

= 0.

Since, f(2) = 0, thus (x - 2) is factor of f(x).

Dividing, f(x) by (x - 2), we get :

$\begin{array}{l} \phantom{x - ]3)}{2x^2 + 5x - 3} \\ x - 2\overline{\smash{\big)}2x^3 + x^2 - 13x + 6} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}4x^2} \\ \phantom{{x - 2}x^,,,3-}5x^2 - 13x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}5x^2 \underset{+}{-}10x} \\ \phantom{{x - 2]euo[ki]}x^3o.\space}{-3x + 6} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{+}{-}3x\underset{-}{+}6} \\ \phantom{{x - 2}{x^,jo-k2x^2\space}{-9x}}\times \end{array}$

∴ 2x³ + x² - 13x + 6 = (x - 2)(2x² + 5x - 3)

= (x - 2)(2x² + 6x - x - 3)

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

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

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

Using factor theorem, factorize the following:

2x³ − x² − 13x − 6

Answer

Let, f(x) = 2x³ − x² − 13x − 6.

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

f(3) = 2(3)³ − (3)² − 13(3) − 6

= 2(27) − 9 − 39 − 6

= 54 − 9 − 39 − 6

= 0.

Since, f(3) = 0, thus (x − 3) is factor of f(x).

Dividing, f(x) by (x − 3), we get :

$\begin{array}{l} \phantom{x - ]3)}{2x^2 + 5x + 2} \\ x - 3\overline{\smash{\big)}2x^3 - x^2 - 13x - 6} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}6x^2} \\ \phantom{{x - 2}x^,,,3-}5x^2 - 13x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}5x^2 \underset{+}{-}15x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{2x - 6} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}2x\underset{+}{-}6} \\ \phantom{{x - 2}{x^,jo-k2x^2\space}{-9x}}\times \end{array}$

∴ 2x³ − x² − 13x − 6 = (x − 3)(2x² + 5x + 2)

= (x − 3)(2x² + 4x + x + 2)

= (x − 3)[2(x + 2) + 1(x + 2)]

= (x − 3)(2x + 1)(x + 2).

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

If (x - 2) is a factor of (x³ + 2x² - kx + 10), find the value of k. Hence, determine whether (x + 5) is also a factor of the given expression.

Answer

Let, f(x) = x³ + 2x² - kx + 10.

Since, x - 2 is factor of f(x), thus f(2) = 0.

∴ (2)³ + 2(2)² - k(2) + 10 = 0

⇒ 8 + 2(4) - 2k + 10 = 0

⇒ 8 + 8 - 2k + 10 = 0

⇒ 26 - 2k = 0

⇒ 2k = 26

⇒ k = $\dfrac{26}{2}$

⇒ k = 13.

f(x) = x³ + 2x² - 13x + 10

⇒ x + 5 = 0

⇒ x = -5.

f(-5) = (-5)³ + 2(-5)² - 13(-5) + 10

= -125 + 2(25) + 65 + 10

= -125 + 50 + 65 + 10

= 125 - 125

= 0.

Since f(−5) = 0, thus (x + 5) is a factor of f(x).

Hence, value of k = 13 and x - 5 is factor of x³ + 2x² - 13x + 10.

Using the remainder and factor theorems, factorize the polynomial.

x³ + 10x² - 37x + 26

Answer

Let, f(x) = x³ + 10x² - 37x + 26.

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

f(1) = (1)³ + 10(1)² - 37(1) + 26

= 1 + 10 - 37(1) + 26

= 37 - 37

= 0.

Since, f(1) = 0, thus (x − 1) is a factor of f(x).

Dividing, x³ + 10x² - 37x + 26 by (x - 1), we get :

$\begin{array}{l} \phantom{x - ]3)}{x^2 + 11x - 26} \\ x - 1\overline{\smash{\big)}x^3 + 10x^2 - 37x + 26} \\ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-}x^2} \\ \phantom{{x - 2}x^,,,3-}11x^2 - 37x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}11x^2 \underset{+}{-}11x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{-26x + 26} \\ \phantom{{x - 2}x^3o;llk]lmttk\space}\underline{\underset{+}{-}26x\underset{-}{+}26} \\ \phantom{{x - 2}{x^,jo-k2x^mm2\space}{-9x}}\times \end{array}$

∴ x³ + 10x² - 37x + 26 = (x - 1)(x² + 11x - 26)

= (x - 1)(x² + 13x - 2x - 26)

= (x - 1)[x(x + 13) - 2(x + 13)]

= (x - 1)(x + 13)(x - 2).

Hence, x³ + 10x² - 37x + 26 = (x - 1)(x + 13)(x - 2).

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

Let f(x) = 2x³ + ax² + bx - 14

Since, (x − 2) is a factor of f(x) then f(2) = 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

⇒ 4a + 2b = -2

⇒ 2(2a + b) = -2

⇒ 2a + b = $\dfrac{-2}{2}$

⇒ 2a + b = -1 ....(1)

On dividing f(x) by (x − 3), the remainder is 52,

By remainder theorem,

⇒ f(3) = 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 = 52 - 40

⇒ 9a + 3b = 12

⇒ 3(3a + b) = 12

⇒ 3a + b = $\dfrac{12}{3}$

⇒ 3a + b = 4 ....(2)

Subtracting equation (1) from (2), we get :

⇒ 3a + b - (2a + b) = 4 - (-1)

⇒ a = 4 + 1

⇒ a = 5.

Substituting a = 5 in equation (1), we get :

⇒ 2(5) + b = -1

⇒ 10 + b = -1

⇒ b = -1 - 10

⇒ b = -11.

Hence, the value of a = 5 and b = -11.

The polynomial 3x³ + 8x² - 15x + k has (x - 1) as a factor. Find the value of k. Hence factorize the resulting polynomial completely.

Answer

⇒ x - 1 = 0

⇒ x = 1.

Given, (x - 1) is a factor of 3x³ + 8x² - 15x + k.

Thus, on substituting x = 1 in 3x³ + 8x² - 15x + k, the remainder will be zero.

⇒ 3.(1)³ + 8.(1)² - 15(1) + k = 0

⇒ 3.1 + 8.1 - 15 + k = 0

⇒ 3 + 8 - 15 + k = 0

⇒ 11 - 15 + k = 0

⇒ k - 4 = 0

⇒ k = 4.

Polynomial = 3x³ + 8x² - 15x + 4

On dividing (3x³ + 8x² - 15x + 4) by (x - 1), we get :

$\begin{array}{l} \phantom{x - 1)}{\quad 3x^2 + 11x - 4} \\ x - 1\overline{\smash{\big)}\quad 3x^3 + 8x^2 - 15x + 4} \\ \phantom{x - 1)}\phantom{)}\underline{\underset{-}{+}3x^3 \underset{+}{-}3x^2} \\ \phantom{{x - 1}31x^3-2}11x^2 - 15x \\ \phantom{{x - 1)}x^3-2}\underline{\underset{-}{+}11x^2 \underset{+}{-} 11x} \\ \phantom{{x - 1)}31x^3-2+1}-4x + 4 \\ \phantom{{x - 1)}31x^3-2+11}\underline{\underset{+}{-}4x \underset{-}{+} 4} \\ \phantom{{x - 1)}31x^3-2+11+1}\times \end{array}$

⇒ 3x³ + 8x² - 15x + 4 = (x - 1)(3x² + 11x - 4)

= (x - 1)[3x² + 12x - x - 4]

= (x - 1)[3x(x + 4) - 1(x + 4)]

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

Hence, 3x³ + 8x² - 15x + 4 = (x - 1)(3x - 1)(x + 4).

It is given that (x − 2) is a factor of polynomial 2x³ − 7x² + kx − 2.

Find:

(i) the value of ‘k’.

(ii) hence, factorise the resulting polynomial completely.

Answer

(i) Since (x − 2) is a factor of 2x³ − 7x² + kx − 2.

Thus, on substituting x = 2, in 2x³ − 7x² + kx − 2, the remainder will be equal to zero.

⇒ 2(2)³ − 7(2)² + k(2) − 2 = 0

⇒ 16 − 28 + 2k − 2 = 0

⇒ −14 + 2k = 0

⇒ 2k = 14

⇒ k = $\dfrac{14}{2}$

⇒ k = 7.

Hence, k = 7.

(ii) Substituting k = 7 in 2x³ − 7x² + kx − 2, we get :

Polynomial : 2x³ − 7x² + 7x − 2.

Dividing 2x³ − 7x² + 7x − 2 by x - 2, we get :

$\begin{array}{l} \phantom{x - 31}{\quad2x^2 - 3x + 1} \\ x - 2\overline{\smash{\big)}\quad 2x^3 − 7x^2 + 7x − 2} \\ \phantom{x^2 + 4}\phantom(\underline{\underset{-}{+}2x^3 \underset{+}{-}4x^2} \\ \phantom{x^2 + 3x - 7)} - 3x^2 + 7x \\ \phantom{x^2 + 3x - 5))}\underline{\underset{+}{-}3x^2 \underset{-}{+}6x} \\ \phantom{x^2 + 3x - 5) + 24x ())}x - 2 \\ \phantom{{x^2 + 3x - 5) + 24 +)}}\underline{\underset{-}{+}x \underset{+}{-}2} \\ \phantom{{x^2 + 3x - 54)} + 2x + 7} \times \end{array}$

⇒ 2x³ − 7x² + 7x − 2 = (x - 2)(2x² - 3x + 1)

⇒ 2x³ − 7x² + 7x − 2 = (x - 2)[2x² - 2x - x + 1]

⇒ 2x³ − 7x² + 7x − 2 = (x - 2)[2x(x - 1) - 1(x - 1)]

⇒ 2x³ − 7x² + 7x − 2 = (x - 2)(2x - 1)(x - 1).

Hence, 2x³ − 7x² + 7x − 2 = (x - 2)(2x - 1)(x - 1).