# How do you factor the expression x^2-x-36?

Jan 5, 2016

Use the quadratic formula to find:

${x}^{2} - x - 36 = \left(x - \frac{1 - \sqrt{145}}{2}\right) \left(x - \frac{1 + \sqrt{145}}{2}\right)$

#### Explanation:

You would like to find a pair of factors of $36$ that differ by $1$. Unfortunately, no such pair exists, so this polynomial will not factor with integer coefficients.

Let's check the discriminant: ${x}^{2} - x - 36$ is in the form $a x + b x + c$ with $a = 1$, $b = - 1$ and $c = - 36$. This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - \left(4 \cdot 1 \cdot - 36\right) = 1 + 144 = 145$

Since this is positive but not a perfect square, the quadratic has factors with Real irrational coefficients.

We can find these factors using the quadratic formula. The zeros of the quadratic are given by:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{1 \pm \sqrt{145}}{2}$

That is ${x}_{1} = \frac{1 - \sqrt{145}}{2}$ and ${x}_{2} = \frac{1 + \sqrt{145}}{2}$

So:

${x}^{2} - x - 36 = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) = \left(x - \frac{1 - \sqrt{145}}{2}\right) \left(x - \frac{1 + \sqrt{145}}{2}\right)$