# How do you factor x^3+x^2-7x-3?

Aug 18, 2016

${x}^{3} + {x}^{2} - 7 x - 3 = \left(x + 3\right) \left(x - 1 - \sqrt{2}\right) \left(x - 1 + \sqrt{2}\right)$

#### Explanation:

$f \left(x\right) = {x}^{3} + {x}^{2} - 7 x - 3$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 3$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means tha the only possible rational zeros are:

$\pm 1 , \pm 3$

We find:

$f \left(- 3\right) = - 27 + 9 + 21 - 3 = 0$

So $x = - 3$ is a zero and $\left(x + 3\right)$ a factor:

${x}^{3} + {x}^{2} - 7 x - 3 = \left(x + 3\right) \left({x}^{2} - 2 x - 1\right)$

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x - 1\right)$ and $b = \sqrt{2}$ as follows:

${x}^{2} - 2 x - 1$

$= {x}^{2} - 2 x + 1 - 2$

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

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

Putting it all together:

${x}^{3} + {x}^{2} - 7 x - 3 = \left(x + 3\right) \left(x - 1 - \sqrt{2}\right) \left(x - 1 + \sqrt{2}\right)$