Factorisation — Chapter Test

Find the remainder when 2x³ - 3x² + 4x + 7 is divided by

(i) x - 2

(ii) x + 3

(iii) 2x + 1

Answer

(i) By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

f(x) = 2x³ - 3x² + 4x + 7

∴ On dividing f(x) by x - 2, Remainder = f(2)

$f(2) = 2(2)^3 - 3(2)^2 + 4(2) + 7 \\[0.5em] = 16 - 12 + 8 + 7 \\[0.5em] = 19$

Hence, the value of remainder is 19.

(ii) By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

f(x) = 2x³ - 3x² + 4x + 7

∴ On dividing f(x) by (x + 3) or (x - (-3)), Remainder = f(-3)

$f(-3) = 2(-3)^3 - 3(-3)^2 + 4(-3) + 7 \\[0.5em] = - 54 - 27 - 12 + 7 \\[0.5em] = -86$

Hence, the value of remainder is -86.

(iii) By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

f(x) = 2x³ - 3x² + 4x + 7

∴ On dividing f(x) by (2x + 1) or $2(x - \big(-\dfrac{1}{2}\big))$, Remainder = f$\big(-\dfrac{1}{2}\big)$

$f(-\dfrac{1}{2}) = 2(-\dfrac{1}{2})^3 - 3(-\dfrac{1}{2})^2 + 4(-\dfrac{1}{2}) + 7 \\[1em] = 2(-\dfrac{1}{8}) - 3(\dfrac{1}{4}) - 2 + 7 \\[1em] = -\dfrac{1}{4} - \dfrac{3}{4} + 5 \\[1em] = \dfrac{-1 -3 + 20}{4} \\[1em] = \dfrac{16}{4} \\[1em] = 4$

Hence, the value of remainder is 4.

When 2x³ - 9x² + 10x - p is divided by (x + 1), the remainder is -24. Find the value of p.

Answer

By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

f(x) = 2x³ - 9x² + 10x - p

∴ On dividing f(x) by (x + 1) or (x - (-1)), Remainder = f(-1)

$\therefore f(-1) = 2(-1)^3 - 9(-1)^2 + 10(-1) - p \\[0.5em] = -2 - 9 - 10 - p \\[0.5em] = -21 - p$

Given, remainder = -24

∴ -21 - p = -24

$\Rightarrow p = 24 - 21 \\[0.5em] p = 3.$

Hence, the value of p is 3.

If (2x - 3) is a factor of 6x² + x + a, find the value of a. With this value of a, factorise the given expression.

Answer

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

f(x) = 6x² + x + a

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

$\therefore 6\Big(\dfrac{3}{2}\Big)^2 + \Big(\dfrac{3}{2}\Big) + a = 0 \\[1em] \Rightarrow 6\Big(\dfrac{9}{4}\Big) + \dfrac{3}{2} + a = 0 \\[1em] \Rightarrow \dfrac{27}{2} + \dfrac{3}{2} + a = 0 \\[1em] \Rightarrow \dfrac{30}{2} + a = 0 \\[1em] \Rightarrow 15 + a = 0 \\[1em] a = -15.$

Putting, a = -15 in f(x) we get,

f(x) = 6x² + x - 15

$\Rightarrow 6x^2 + 10x - 9x - 15 \\[0.5em] \Rightarrow 2x(3x + 5) - 3(3x + 5) \\[0.5em] (2x - 3)(3x + 5)$

Hence, the value of a is -15;
6x² + x - 15 = (2x - 3)(3x + 5).

When 3x² - 5x + p is divided by (x - 2), the remainder is 3. Find the value of p. Also factorise the polynomial 3x² - 5x + p - 3.

Answer

By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

f(x) = 3x² - 5x + p

∴ On dividing f(x) by (x - 2), Remainder = f(2)

Given, Remainder = 3

∴ f(2) = 3

$\Rightarrow 3(2)^2 - 5(2) + p = 3 \\[0.5em] \Rightarrow 12 - 10 + p = 3 \\[0.5em] \Rightarrow p + 2 = 3 \\[0.5em] \Rightarrow p = 3 - 2 \\[0.5em] p = 1$

Putting value of p = 1 in 3x² - 5x + p - 3,

$3x^2 - 5x + 1 - 3 \\[0.5em] = 3x^2 - 5x - 2 \\[0.5em] = 3x^2 - 6x + x - 2 \\[0.5em] = 3x(x - 2) + 1(x - 2) \\[0.5em] (3x + 1)(x - 2)$

Hence, the value of p is 1 and the factors are (3x + 1) and (x - 2).

Prove that (5x + 4) is a factor of 5x³ + 4x² - 5x - 4. Hence, factorise the given polynomial completely.

Answer

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

f(x) = 5x³ + 4x² - 5x - 4

Given, (5x + 4) or 5(x - (-$\dfrac{4}{5}$)) is a factor of f(x) hence, let's find f(-$\dfrac{4}{5}$).

$\therefore f(-\dfrac{4}{5}) = 5\big(-\dfrac{4}{5}\big)^3 + 4\big(-\dfrac{4}{5}\big)^2 -5\big(-\dfrac{4}{5}\big) - 4 \\[1em] = 5\big(-\dfrac{64}{125}\big) + 4\big(\dfrac{16}{25}\big) + 4 - 4 \\[1em] = -\dfrac{64}{25} + \dfrac{64}{25} \\[1em] = 0$

Since, f(-$\dfrac{4}{5}$) = 0, hence (5x + 4) is a factor of f(x).

Now, factorising the equation 5x³ + 4x² - 5x - 4

$\Rightarrow x^2(5x + 4) - 1(5x + 4) \\[0.5em] \Rightarrow (x^2 - 1)(5x + 4) \\[0.5em] \Rightarrow (x^2 - (1)^2)(5x + 4) \\[0.5em] \Rightarrow (x - 1)(x + 1)(5x + 4)$

Hence, 5x³ + 4x² - 5x - 4 = (x - 1)(x + 1)(5x + 4).

Use factor theorem to factorise the following polynomials completely :

(i) 4x³ + 4x² - 9x - 9

(ii) x³ - 19x - 30

(iii) 2x³ - x² - 13x - 6

Answer

(i) f(x) = 4x³ + 4x² - 9x - 9

Let x = -1, substituting the value of x in f(x),

$f(-1) = 4(-1)^3 + 4(-1)^2 - 9(-1) - 9 \\[0.5em] = -4 + 4 + 9 - 9 \\[0.5em] = 0$

Since, f(-1) = 0 hence, (x + 1) is a factor of 4x³ + 4x² - 9x - 9.

On dividing, 4x³ + 4x² - 9x - 9 by (x + 1),

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

we get (4x² - 9) as quotient and remainder = 0.

$\therefore 4x^3 + 4x^2 - 9x - 9 = (x + 1)(4x^2 - 9) \\[0.5em] = (x + 1)((2x)^2 - (3)^2) \\[0.5em] = (x + 1)(2x - 3)(2x + 3)$

Hence, 4x³ + 4x² - 9x - 9 = (x + 1) (2x - 3)(2x + 3).

(ii) f(x) = x³ - 19x - 30

Let x = -2, substituting the value of x in f(x),

$f(-2) = (-2)^3 - 19(-2) - 30 \\[0.5em] = -8 + 38 - 30 \\[0.5em] = 0$

Since, f(-2) = 0 hence, (x + 2) is a factor of x³ - 19x - 30.

On dividing, x³ - 19x - 30 by (x + 2),

$\begin{array}{l} \phantom{x + 2)}{x^2 - 2x - 15} \\ x + 2\overline{\smash{\big)}x^3 - 19x - 30} \\ \phantom{x + 2}\underline{\underset{-}{ }x^3 \underset{-}{+} 2x^2} \\ \phantom{{x + 2}{x^3+}}-2x^2 - 19x \\ \phantom{{x + 2}x^3+}\underline{\underset{+}{-}2x^2 \underset{+}{-} 4x} \\ \phantom{{x + 2}{-x^3+2x^2}}-15x - 30 \\ \phantom{{x + 2}{-x^3+2x^2+}}\underline{\underset{+}{-}15x \underset{+}{-} 30} \\ \phantom{{x + 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get x² - 2x - 15 as quotient and remainder = 0.

$\therefore x^3 - 19x - 30 = (x + 2)(x^2 - 2x - 15) \\[0.5em] = (x + 2)(x^2 - 5x + 3x - 15) \\[0.5em] = (x + 2)(x(x - 5) + 3(x - 5)) \\[0.5em] = (x + 2)(x + 3)(x - 5)$

Hence, x³ - 19x - 30 = (x + 2) (x + 3)(x - 5).

(iii) f(x) = 2x³ - x² - 13x - 6

Substituting x = -2 in 2x³ - x² - 13x - 6, we get :

⇒ 2(-2)³ - (-2)² - 13(-2) - 6

⇒ 2(-8) - 4 + 26 - 6

⇒ -16 - 4 + 20

⇒ -20 + 20

⇒ 0.

∴ x + 2 is a factor of the polynomial 2x³ - x² - 13x - 6.

Dividing, 2x³ - x² - 13x - 6 by x + 2, we get :

$\begin{array}{l} \phantom{x + 2)}{\quad 2x^2 -5x - 3} \\ x + 2\overline{\smash{\big)}\quad 2x^3 - x^2 - 13x - 6} \\ \phantom{x + 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{-}{+}4x^2} \\ \phantom{{x + 2}x^3-2}-5x^2 - 13x \\ \phantom{{x + 2)}x^3-2}\underline{\underset{+}{-}5x^2 \underset{+}{-} 10x} \\ \phantom{{x + 2)}{x^3-2x^{2}(3)}}-3x - 6 \\ \phantom{{x + 2)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}3x \underset{+}{-} 6} \\ \phantom{{x + 2)}{x^3-2x^{2}(31)}{-2x}}\times \end{array}$

∴ 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).

If x³ - 2x² + px + q has a factor (x + 2) and leaves a remainder 9 when divided by (x + 1), find the values of p and q. With these values of p and q, factorise the given polynomial completely.

Answer

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

f(x) = x³ - 2x² + px + q

Given, (x + 2) or (x - (-2)) is a factor of f(x).

∴ f(-2) = 0

$\Rightarrow (-2)^3 - 2(-2)^2 + p(-2) + q = 0 \\[0.5em] \Rightarrow -8 - 8 - 2p + q = 0 \\[0.5em] \Rightarrow -2p + q = 16 \\[0.5em] q = 2p + 16 \text{ \space (Equation 1)}$

By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

∴ On dividing f(x) by (x + 1) or (x - (-1)), Remainder = f(-1)

Given, Remainder = 9

∴ f(-1) = 9

$\Rightarrow (-1)^3 - 2(-1)^2 + p(-1) + q = 9 \\[0.5em] \Rightarrow -1 - 2 - p + q = 9 \\[0.5em] \Rightarrow -3 - p + q = 9 \\[0.5em] \Rightarrow q - p = 9 + 3 \\[0.5em] \Rightarrow q - p = 12$

Putting value of q = 2p + 16 from equation 1,

$\Rightarrow 2p + 16 - p = 12 \\[0.5em] \Rightarrow p + 16 = 12 \\[0.5em] \Rightarrow p = 12 - 16 \\[0.5em] \Rightarrow p = -4 \\[0.5em] \text{ and } q = 2p + 16 = 2(-4) + 16 = -8 + 16 = 8$

Now putting p = -4 and q = 8 in f(x),

f(x) = x³ - 2x² - 4x + 8

Since, (x + 2) is a factor of f(x), on dividing f(x) by (x + 2),

$\begin{array}{l} \phantom{x + 2)}{x^2 - 4x + 4} \\ x + 2\overline{\smash{\big)}x^3 - 2x^2 - 4x + 8} \\ \phantom{x + 2}\underline{\underset{-}{ }x^3 \underset{-}{+} 2x^2} \\ \phantom{{x + 2}{x^3+}}-4x^2 - 4x \\ \phantom{{x + 2}x^3+}\underline{\underset{+}{-}4x^2 \underset{+}{-} 8x} \\ \phantom{{x + 2}{-x^3+2x^21+}}4x + 8 \\ \phantom{{x + 2}{-x^3+2x^2+}}\underline{\underset{-}{ }4x \underset{-}{+} 8} \\ \phantom{{x + 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get x² - 4x + 4 as quotient and remainder = 0.

$\therefore x^3 - 2x^2 - 4x + 8 = (x + 2)(x^2 - 4x + 4) \\[0.5em] =(x + 2)(x^2 - 2 \times 2 \times x + 2^2) \\[0.5em] = (x + 2)(x - 2)^2$

Hence, value of p = -4 and q = 8;
x³ - 2x² - 4x + 8 = (x + 2) (x - 2)².

If (x + 3) and (x - 4) are factors of x³ + ax² - bx + 24, find the values of a and b. With these values of a and b, factorise the given expression.

Answer

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

f(x) = x³ + ax² - bx + 24

Given, (x + 3) or (x - (-3) and (x - 4) are factors of f(x)

∴ f(-3) = 0 and f(4) = 0.

For, f(-3) = 0

$\Rightarrow (-3)^3 + a(-3)^2 - b(-3) + 24 = 0 \\[0.5em] \Rightarrow -27 + 9a + 3b + 24 = 0 \\[0.5em] \Rightarrow 9a + 3b - 3 = 0 \\[0.5em] \Rightarrow 9a + 3b = 3$

On dividing equation by 3,

$\Rightarrow 3a + b = 1 \\[0.5em] b = 1 - 3a \text{ \space (Equation 1) }$

For f(4) = 0

$\Rightarrow (4)^3 + a(4)^2 - b(4) + 24 = 0 \\[0.5em] \Rightarrow 64 + 16a - 4b + 24 = 0 \\[0.5em] \Rightarrow 16a - 4b + 88 = 0 \\[0.5em] \Rightarrow 16a - 4b = -88$

On dividing equation by 4,

$\Rightarrow 4a - b = -22$

Putting value of b = 1 - 3a from equation 1,

$\Rightarrow 4a - 1 + 3a = -22 \\[0.5em] \Rightarrow 7a - 1 = -22 \\[0.5em] \Rightarrow 7a = -22 + 1 \\[0.5em] \Rightarrow 7a = -21 \\[1em] \Rightarrow a = -\dfrac{21}{7} \\[1em] \Rightarrow a = -3 \\[0.5em] \text{ and } b = 1 - 3a = 1 - 3(-3) = 1 + 9 = 10$

Now putting a = -3 and b = 10 in f(x),

f(x) = x³ - 3x² - 10x + 24

Since, (x + 3) and (x - 4) is a factor of f(x), hence (x + 3)(x + 4) is also the factor

$(x + 3)(x - 4) = x^2 + 3x - 4x - 12 \\[0.5em] = x^2 - x - 12$

On dividing, f(x) by x² - x - 12,

$\begin{array}{l} \phantom{x^2 - x - 12)}{x - 2} \\ x^2 - x - 12\overline{\smash{\big)}x^3 - 3x^2 - 10x + 24} \\ \phantom{x^2 - x - 12}\underline{\underset{-}{ }x^3 \underset{+}{-} x^2 \underset{+}{-} 12x} \\ \phantom{{x^2 - x - 12}{x^3+}}-2x^2 + 2x + 24 \\ \phantom{{x^2 - x - 12}x^3+}\underline{\underset{+}{-}2x^2 \underset{-}{+} 2x \underset{-}{+} 24} \\ \phantom{{x^2 - x - 12}{2x^3+}{+2x^2-}}\times \end{array}$

we get (x - 2) as quotient and remainder = 0.

$\therefore x^3 - 3x^2 - 10x + 24 = (x - 2)(x^2 - x - 12). \\[0.5em] = (x - 2)(x^2 - 4x + 3x - 12) \\[0.5em] = (x - 2)(x(x - 4) + 3(x - 4)) \\[0.5em] = (x - 2)(x + 3)(x - 4)$

Hence, value of a = -3 and b = 10;
x³ - 3x² - 10x + 24 = (x - 2)(x + 3)(x - 4).

If (2x + 1) is a factor of both the expressions 2x² - 5x + p and 2x² + 5x + q, find the values of p and q. Hence, find the other factors of both the polynomials.

Answer

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

Let, f(x) = 2x² - 5x + p

Given, (2x + 1) or 2(x - (-$\dfrac{1}{2}$) is a factor of f(x)

$\therefore f(-\dfrac{1}{2}) = 0 \\[1em] \Rightarrow 2\big(-\dfrac{1}{2}\big)^2 - 5\big(-\dfrac{1}{2}\big) + p = 0 \\[1em] \Rightarrow 2\big(\dfrac{1}{4}) + \dfrac{5}{2} + p = 0 \\[1em] \Rightarrow \dfrac{1}{2} + \dfrac{5}{2} + p = 0 \\[1em] \Rightarrow \dfrac{6}{2} + p = 0 \\[1em] \Rightarrow 3 + p = 0 \\[1em] p = -3$

Putting value of p in f(x)

$f(x) = 2x^2 - 5x - 3 \\[0.5em] = 2x^2 - 6x + x - 3 \\[0.5em] = 2x(x - 3) + 1(x - 3) \\[0.5em] = (2x + 1)(x - 3)$

Hence, p = -3 and other factor is (x - 3).

Let, g(x) = 2x² + 5x + q

Given, (2x + 1) or 2(x - (-$\dfrac{1}{2}$) is a factor of g(x)

$\therefore g(-\dfrac{1}{2}) = 0 \\[0.5em] \Rightarrow 2\big(-\dfrac{1}{2}\big)^2 + 5\big(-\dfrac{1}{2}\big) + q = 0 \\[0.5em] \Rightarrow 2\big(\dfrac{1}{4}) - \dfrac{5}{2} + q = 0 \\[0.5em] \Rightarrow \dfrac{1}{2} - \dfrac{5}{2} + q = 0 \\[0.5em] \Rightarrow -\dfrac{4}{2} + q = 0 \\[0.5em] \Rightarrow -2 + q = 0 \\[0.5em] q = 2$

Putting value of q in g(x)

$f(x) = 2x^2 + 5x + 2 \\[0.5em] = 2x^2 + 4x + x + 2 \\[0.5em] = 2x(x + 2) + 1(x + 2) \\[0.5em] = (2x + 1)(x + 2)$

Hence, q = 2 and other factor is (x + 2).

If a polynomial f(x) = x⁴ - 2x³ + 3x² - ax - b leaves remainders 5 and 19 when divided by (x - 1) and (x + 1) respectively, find the values of a and b. Hence, determine the remainder when f(x) is divided by (x - 2).

Answer

By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

f(x) = x⁴ - 2x³ + 3x² - ax - b

∴ On dividing f(x) by (x + 1) or (x - (-1)), Remainder = f(-1)

Given, on dividing by (x + 1) remainder = 19,

∴ f(-1) = 19

$\Rightarrow (-1)^4 - 2(-1)^3 + 3(-1)^2 - a(-1) - b = 19 \\[0.5em] \Rightarrow 1 + 2 + 3 + a - b = 19 \\[0.5em] \Rightarrow a - b + 6 = 19 \\[0.5em] \Rightarrow a - b = 13 \\[0.5em] a = 13 + b \text{ \space (Equation 1) }$

∴ On dividing f(x) by (x - 1), Remainder = f(1)

Given, on dividing by (x - 1) remainder = 5,

∴ f(1) = 5

$\Rightarrow (1)^4 - 2(1)^3 + 3(1)^2 - a(1) - b = 5 \\[0.5em] \Rightarrow 1 - 2 + 3 - a - b = 5 \\[0.5em] \Rightarrow 2 - a - b = 5 \\[0.5em] \Rightarrow -a - b = 3$

Putting value of a = 13 + b from equation 1,

$\Rightarrow -(13 + b) - b = 3 \\[0.5em] \Rightarrow -13 - b - b = 3 \\[0.5em] \Rightarrow -13 - 2b = 3 \\[0.5em] \Rightarrow 2b = -16 \\[0.5em] \Rightarrow b = -\dfrac{16}{2} \\[0.5em] \Rightarrow b = -8 \\[0.5em] \text{ and } a = 13 + b = 13 - 8 = 5$

Putting a = 5 and b = -8 in f(x) we get,

f(x) = x⁴ - 2x³ + 3x² - 5x + 8.

On dividing f(x) by (x - 2), remainder = f(2) by remainder theorem

$f(2) = (2)^4 - 2(2)^3 + 3(2)^2 - 5(2) + 8 \\[0.5em] = 16 - 16 + 12 - 10 + 8 \\[0.5em] = 10.$

Hence, the value of a is 5 and b is -8.
On dividing x⁴ - 2x³ + 3x² - 5x + 8 by (x - 2) the value of remainder is 10.

When a polynomial f(x) is divided by (x - 1), the remainder is 5 and when it is, divided by (x - 2), the remainder is 7. Find the remainder when it is divided by (x - 1)(x - 2).

Answer

By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

Given, when f(x) is divided by (x - 1), remainder = 5

∴ f(1) = 5

Given, when f(x) is divided by (x - 2), remainder = 7

∴ f(2) = 7

Suppose on dividing f(x) by (x - 1)(x - 2),

Quotient = q(x)

Remainder = ax + b

So, f(x) = (x - 1)(x - 2)q(x) + ax + b

Putting x = 1, we get:

$\Rightarrow f(1) = (1 - 1)(1 - 2)q(1) + a(1) + b = 5 \\[0.5em] \Rightarrow 0 + a + b = 5 \\[0.5em] \Rightarrow a + b = 5 \\[0.5em] a = 5 - b \text{ \space (Equation 1)}$

Putting x = 2, we get:

$\Rightarrow f(2) = (2 - 1)(2 - 2)q(2) + a(2) + b = 7 \\[0.5em] \Rightarrow 0 + 2a + b = 7 \\[0.5em] \Rightarrow 2a + b = 7$

Putting value of a from equation 1,

$\Rightarrow 2(5 - b) + b = 7 \\[0.5em] \Rightarrow 10 - 2b + b = 7 \\[0.5em] \Rightarrow b = 10 - 7 \\[0.5em] \Rightarrow b = 3 \\[0.5em] \text{and } a = 5 - b = 5 - 3 = 2.$

Remainder = ax + b = 2x + 3.

∴ The remainder when polynomial is divided by (x - 1)(x - 2) is 2x + 3.

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).

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

(a) Is the student's solution correct stating that (2x + 1) is a factor of the given polynomial ? Give a valid reason for your answer.

(b) Factorize the given polynomial completely.

Answer

⇒ 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 :

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

Since, remainder is equal to zero.

∴ x - $\dfrac{1}{2}$ is factor of polynomial,

⇒ x - $\dfrac{1}{2}$ = 0

⇒ x = $\dfrac{1}{2}$

⇒ 2x = 1

⇒ 2x - 1 is factor of polynomial.

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

$\begin{array}{l} \phantom{2x - 1)}{\quad x^2 + 4x + 3} \\ 2x - 1\overline{\smash{\big)}\quad 2x^3 + 7x^2 + 2x - 3} \\ \phantom{2x - 1)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{+}{-}x^2} \\ \phantom{{2x - 1}+2x^3 + 1}8x^2 + 2x \\ \phantom{{2x - 1)}+2x^31}\underline{\underset{-}{+}8x^2 \underset{+}{-} 4x} \\ \phantom{{2x - 1)}+2x^31+8x^212}6x - 3 \\ \phantom{{2x - 1)}+2x^31+8x^21}\underline{\underset{-}{+}6x \underset{+}{-} 3} \\ \phantom{{2x - 1)}+2x^31+8x^21+6}\times \end{array}$

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).