Question

Find all the zeros of polynomial p (x)= x⁴-2x³-7x²+8x+12. If the zeros are -1 and 2

Answer 1

Explanation:

Since we know that -1 is one of the zeroes, it means that (x + 1) is a factor of p(x).

∴ $\mathsf{p(x)=x^4-2x^3-7x^2+8x+12}$

$\mathsf{=x^3(x+1)-3x^2(x+1)-4x(x+1)+12(x+1)}$

$\mathsf{=(x+1)(x^3-3x^2-4x+12)}$

$\mathsf{=(x+1)[x^2(x-3)-4(x-3)]}$

$\mathsf{=(x+1)(x-3)(x^2-4)}$

$\mathsf{=(x+1)(x+2)(x-2)(x-3)}$ [∵$\mathsf{a^2-b^2=(a+b)(a-b)}$]

Hence, the zeroes of p(x) are -1, -2, 2 and 3.

Answer: -1, -2, 2, 3

I hope this helps. :D