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.

Question

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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). 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).

Mathematics

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.

Factorisation

Related Questions

In the given diagram, O is the centre of the circle and the tangent DE touches the circle at B. If ∠ADB = 32°. Find the values of x and y.

Mathematics

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.

Factorisation

Related Questions

In the given diagram, O is the centre of the circle and the tangent DE touches the circle at B. If ∠ADB = 32°. Find the values of x and y.

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.