# How do you factor 6x^2y + 36xy - 66y?

May 19, 2015

All the terms are multiples of $6 y$, so we find:

$6 {x}^{2} y + 36 x y - 66 y = 6 y \left({x}^{2} + 6 x - 11\right)$

${x}^{2} + 6 x - 11$ is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 6$ and $c = - 11$, which has discriminant given by the formula:

$\Delta = {b}^{2} - 4 a c = {6}^{2} - \left(4 \times 1 \times - 11\right) = 36 + 44 = 80$

This is positive, but not a perfect square. So the zeros of ${x}^{2} + 6 x - 11$ are not rational and its linear factors will have irrational coefficients.

The roots of ${x}^{2} + 6 x - 11 = 0$ are given by the formula:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 6 \pm \sqrt{80}}{2}$

$= \frac{- 6 \pm 4 \sqrt{5}}{2}$

$= - 3 \pm 2 \sqrt{5}$

This gives us factors $\left(x + 3 + 2 \sqrt{5}\right)$ and $\left(x + 3 - 2 \sqrt{5}\right)$

So the full factorization (if we allow irrational coefficients) is:

$6 {x}^{2} y + 36 x y - 66 y = 6 y \left(x + 3 + 2 \sqrt{5}\right) \left(x + 3 - 2 \sqrt{5}\right)$