# How do you determine the binomial factors of x^3+4x^2-x-4?

Dec 14, 2016

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

#### Explanation:

It is apparent from the given polynomial ${x}^{3} + 4 {x}^{2} - x - 4$, that if ${x}^{2}$ is taken from first two terms, we have $x + 4$ left out and this is also left out if take out $- 1$ from remaining two terms.

Hence ${x}^{3} + 4 {x}^{2} - x - 4$

= ${x}^{2} \left(x + 4\right) - 1 \left(x + 4\right)$

= $\left({x}^{2} - 1\right) \left(x + 4\right)$

= $\left({x}^{2} + x - x - 1\right) \left(x + 4\right)$

= $\left(x \left(x + 1\right) - 1 \left(x + 1\right)\right) \left(x + 4\right)$

= $\left(x - 1\right) \left(x + 1\right) \left(x + 4\right)$