# How do you factor 20w^2-17w-63?

Feb 7, 2017

$20 {w}^{2} - 17 w - 63 = \left(5 w + 7\right) \left(4 w - 9\right)$

#### Explanation:

Given:

$20 {w}^{2} - 17 w - 63$

This is in the form:

$a {w}^{2} + b w + c$

with $a = 20$, $b = - 17$ and $c = - 63$

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

$\Delta = {b}^{2} - 4 a c = {\left(- 17\right)}^{2} - 4 \left(20\right) \left(- 63\right) = 289 + 5040 = 5329 = {73}^{2}$

Since $\Delta$ is a perfect square, this quadratic will factor using integer coefficients.

Use an AC method:

Find a pair of factors of $A C = 20 \cdot 63 = 1260$ which differ by $B = 17$.

The prime factorisation of $1260$ is:

$1260 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 7$

With a bit of reasoning and trial and error, we find the pair $45 , 28$ works.

Use this pair to split the middle term and factor by grouping:

$20 {w}^{2} - 17 w - 63 = \left(20 {w}^{2} - 45 w\right) + \left(28 w - 63\right)$

$\textcolor{w h i t e}{20 {w}^{2} - 17 w - 63} = 5 w \left(4 w - 9\right) + 7 \left(4 w - 9\right)$

$\textcolor{w h i t e}{20 {w}^{2} - 17 w - 63} = \left(5 w + 7\right) \left(4 w - 9\right)$