How do you factor #100+4x^2-16y-40x#?

1 Answer
George C.
Apr 21, 2017

If the #-16y# should have been #-16y^2#, then we find:

#100+4x^2-16y^2-40x = 4(x-2y-5)(x+2y-5)#

Explanation:

For the record: I think the question should have specified #-16y^2# rather than #-16y#. Here's what happens if that is the case:

The difference of squares identity can be written:

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

Use this with #a=(x-5)# and #b=2y# as follows:

#100+4x^2-16y^2-40x = 4(x^2-10x+25-4y^2)#

#color(white)(100+4x^2-16y^2-40x) = 4(x^2-2x(5)+5^2-4y^2)#

#color(white)(100+4x^2-16y^2-40x) = 4((x-5)^2-(2y)^2)#

#color(white)(100+4x^2-16y^2-40x) = 4((x-5)-2y)((x-5)+2y)#

#color(white)(100+4x^2-16y^2-40x) = 4(x-2y-5)(x+2y-5)#

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