# How do you determine the binomial factors of x^3-5x^2+2x+8?

Dec 28, 2016

${x}^{3} - 5 {x}^{2} + 2 x + 8 = \left(x + 1\right) \left(x - 4\right) \left(x - 2\right)$

#### Explanation:

Given:

${x}^{3} - 5 {x}^{2} + 2 x + 8$

Notice that if we reverse the signs on the coefficients of the terms with odd degree then the sum of the coefficients is zero.

That is:

$- 1 - 5 - 2 + 8 = 0$

Hence $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

${x}^{3} - 5 {x}^{2} + 2 x + 8 = \left(x + 1\right) \left({x}^{2} - 6 x + 8\right)$

To factor the remaining quadratic note that $6 = 4 + 2$ and $8 = 4 \cdot 2$, so we find:

${x}^{2} - 6 x + 8 = \left(x - 4\right) \left(x - 2\right)$

Putting it all together:

${x}^{3} - 5 {x}^{2} + 2 x + 8 = \left(x + 1\right) \left(x - 4\right) \left(x - 2\right)$
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