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.
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(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).