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)#