# How do you find the zeros of the polynomial function with equation f(x)=x^3+8x^2-x-8?

Mar 28, 2016

x = -8 , x = ± 1

#### Explanation:

The zeros are the values of x , which make the function equal zero.

ie ${x}^{3} + 8 {x}^{2} - x - 8 = 0$

To solve this equation for x , require to factorise it.

Consider the function 'split' into 2 pairs of terms.

hence $\left[{x}^{3} + 8 {x}^{2}\right] + \left[- x - 8\right]$

now factor each pair

thus ${x}^{2} \left(x + 8\right) - 1 \left(x + 8\right) = \left(x + 8\right) \left({x}^{2} - 1\right)$

$\Rightarrow \left(x + 8\right) \left({x}^{2} - 1\right) = 0$

solve x + 8 = 0 → x = -8

solve  x^2 -1 = 0 → (x+1)(x-1) = 0 → x = ± 1