# How do you factor x^3+9x^2+15x-25?

Nov 11, 2016

${x}^{3} + 9 {x}^{2} + 15 x - 25 = \left(x - 1\right) {\left(x + 5\right)}^{2}$

#### Explanation:

Given:

${x}^{3} + 9 {x}^{2} + 15 x - 25$

First notice that the sum of the coefficients is $0$. That is:

$1 + 9 + 15 - 25 = 0$

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

${x}^{3} + 9 {x}^{2} + 15 x - 25 = \left(x - 1\right) \left({x}^{2} + 10 x + 25\right)$

To factor the remaining quadratic note that both ${x}^{2}$ and $25 = {5}^{2}$ are perfect squares, with $10 x = 2 \left(5\right) x$. So this quadratic is a perfect square trinomial:

${x}^{2} + 10 x + 25 = {x}^{2} + 2 \left(5\right) x + {5}^{2} = {\left(x + 5\right)}^{2}$

Putting it all together:

${x}^{3} + 9 {x}^{2} + 15 x - 25 = \left(x - 1\right) {\left(x + 5\right)}^{2}$