# How do you factor 4w^2 -13w -27?

May 23, 2015

$4 {w}^{2} - 13 w - 27$ is of the form $a {w}^{2} + b w + c$, with $a = 4$, $b = - 13$ and $c = - 27$.

This has discriminant given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 13\right)}^{2} - \left(4 \times 4 \times - 27\right)$

$= 169 + 432 = 601$

Though positive, this is not a perfect square. So the values of $w$ for which $4 {w}^{2} - 13 w - 27 = 0$ are real, but irrational. These roots are:

$w = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{13 \pm \sqrt{601}}{8}$

Hence:

$4 {w}^{2} - 13 w - 27$

=4(w-(13+sqrt(601))/8))(w-(13-sqrt(601))/8))

=(2w-(13+sqrt(601))/4)(2w-(13-sqrt(601))/4))

If the original problem were to factor $4 {w}^{2} - 12 w - 27$ the result would have been the much simpler:

$4 {w}^{2} - 12 w - 27 = \left(2 w - 9\right) \left(2 w + 3\right)$

which you could spot using the rational root theorem and a little trial and error.