How do you factor #x² - 4x - 45#?

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Answer 1 · 2015-07-06T21:34:58

Find two factors of #45# whose difference is #4#. The numbers #9# and #5# work.

Hence #x^2-4x-45 = (x-9)(x+5)#

#(x-a)(x+b) = x^2-(a-b)x-ab#

So given #x^2-4x-45#, look for #a# and #b#

such that #ab=45# and #a-b = 4#.

Answer 2 · 2015-07-06T21:51:45

Alternatively, complete the square to find:

#x^2-4x-45 = (x-2)^2-7^2 =#

#(x-2-7)(x-2+7) = (x-9)(x+5)#

#x^2-4x-45#

#=x^2-4x+4-4-45#

#=(x-2)^2-49#

#=(x-2)^2-7^2#

This is a difference of squares, so we can use the identity:

#a^2-b^2 = (a-b)(a+b)#

Let #a=x-2# and #b=7#

Then

#(x-2)^2-7^2#

#=(x-2-7)(x-2+7) = (x-9)(x+5)#