How do you factor #x^2+ 11x = 180#?

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Answer 1 · 2015-05-24T09:01:01

First subtract #180# from both sides to get:

#x^2+11x-180 = 0#

Noticing that the coefficient of #x^2# is #1# and the signs of the other two coefficients, this may factorize as:

#(x+a)(x-b) = x^2+(a-b)x-ab# with #a > 0#, #b > 0#, #a-b = 11# and #ab = 180#.

#180 = 20 xx 9# and #11 = 20 - 9#

So we can let #a=20# and #b=9# to get:

#x^2+11x-180 = (x+20)(x-9)#

Hence #x^2+11x=180# has solutions #x=-20# and #x=9#