Question

Factorise: xcube+3xsquare-7x-6

Answer 1

Step-by-step explanation:

x3+3.x2.1+3.x.12+13-8

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

=(x+1-2){(x+1)2+2.(x+1)+22}

=(x-1)(x2+2x+1+2x+2+4)

=(x-1)(x2+4x+7)

Answer 2

Assume, p(x) = x^3 + 3x^2 - 7x - 6.

Factors of the constant term (6) = ± 1, ± 2, ± 3, ± 4, ± 6, ± 5.

By putting 2 as x in p(x) we got that p(2) is a zero.

{If α is a zero of a polynomial, then p(α) = 0.}

So, (x - 2) is a factor.

Now, by performing synthetic division process, we got that when dividend = p(x) and divisior = (x - 2), we get remainder = 0 and quotient = x^2 + 5x + 3.

Therefore, (x - 2)(x^2 + 5x + 3) is the factorised form.

{x^2 + 5x + 3 cannot be factored.}