The polynomial kx³ + 3x² − 11x − 6 when divided by (x + 1), leaves a remainder of 6.

(a) Find the value of k.

(b) Using the value of k factorise completely the polynomial kx³ + 3x² − 11x − 6.

(a) Using remainder theorem,

If polynomial P(x) is divided by (x - a), the remainder is P(a).

x + 1 = 0

x = -1.

Given,

The polynomial kx³ + 3x² − 11x − 6 when divided by (x + 1), leaves a remainder of 6.

Thus, on substituting value x = -1 in kx³ + 3x² − 11x − 6, we get :

⇒ k(-1)³ + 3(-1)² − 11(-1) − 6 = 6

⇒ -k + 3 + 11 − 6 = 6

⇒ -k + 8 = 6

⇒ -k = 6 - 8

⇒ k = 2.

Hence, the value of k = 2.

(b) P(x) = 2x³ + 3x² − 11x − 6

Substituting value x = 2 in P(x), we get :

⇒ 2(2)³ + 3(2)² − 11(2) − 6

⇒ 2(8) + 3(4) − 22 − 6

⇒ 16 + 12 − 22 − 6

⇒ 28 − 28

⇒ 0.

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

$\begin{array}{l} \phantom{x - 2)}{2x^2 + 7x + 3} \ x - 2\overline{\smash{\big)}2x^3 + 3x^2 − 11x − 6} \ \phantom{x - 2}\phantom{}\underline{\underset{-}{}2x^3 \underset{+}{-}4x^2} \ \phantom{{x - 2}x^3,.-2}7x^2 - 11x \ \phantom{{x - 2)}x^3-2}\underline{\underset{-}{}7x^2 \underset{+}{-} 14x} \ \phantom{{x - 2)}{x^3-2x^{2}(3-.)}}3x - 6 \ \phantom{{x - 2)}{x^3-2x^{2}(31),}}\underline{\underset{-}{}3x \underset{+}{-} 6} \ \phantom{{x - 2)}{x^3-2x^{2}(31)}{-2x}}\times \end{array}$

∴ 2x³ + 3x² − 11x − 6 = (x - 2)(2x² + 7x + 3)

= (x - 2)(2x² + 6x + x + 3)

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

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

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