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

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Answer 1 · 2017-02-07T00:32:10

$20w^2-17w-63 = (5w+7)(4w-9)$

Given:

$20w^2-17w-63$

This is in the form:

$aw^2+bw+c$

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

It has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = (-17)^2-4(20)(-63) = 289+5040 = 5329 = 73^2$

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

Use an AC method:

Find a pair of factors of $AC=20*63 = 1260$ which differ by $B=17$ .

The prime factorisation of $1260$ is:

$1260 = 2*2*3*3*5*7$

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:

$20w^2-17w-63 = (20w^2-45w)+(28w-63)$

$color(white)(20w^2-17w-63) = 5w(4w-9)+7(4w-9)$

$color(white)(20w^2-17w-63) = (5w+7)(4w-9)$

How do you factor 20w^2-17w-6320w^2-17w-63?