# Is it possible to factor y=x^2 - 9x+36? If so, what are the factors?

Jan 7, 2016

Yes, but only with Complex coefficients:

${x}^{2} - 9 x + 36 = \left(x - \frac{9}{2} - \frac{3 \sqrt{7}}{2} i\right) \left(x - \frac{9}{2} + \frac{3 \sqrt{7}}{2} i\right)$

#### Explanation:

$y = {x}^{2} - 9 x + 36$ is in the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 9$ and $c = 36$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 9\right)}^{2} - \left(4 \cdot 1 \cdot 36\right) = 81 - 144 = - 63$

Since this is negative, this quadratic has no factors with Real coefficients.

If you still want to factor it, use the quadratic formula to find the Complex zeros:

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

$= \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{9 \pm \sqrt{- 63}}{2}$

$= \frac{9 \pm \sqrt{63} i}{2}$

$= \frac{9 \pm \sqrt{{3}^{2} \cdot 7} i}{2}$

$= \frac{9 \pm 3 \sqrt{7} i}{2}$

So:

${x}^{2} - 9 x + 36 = \left(x - \frac{9}{2} - \frac{3 \sqrt{7}}{2} i\right) \left(x - \frac{9}{2} + \frac{3 \sqrt{7}}{2} i\right)$