How do you factor #200x^2 + 800x +800#?

1 Answer
May 12, 2016

Answer:

#200x^2+800x+800 = 200(x+2)^2#

Explanation:

All of the terms are divisible by #200#, so separate that out first:

#200x^2+800x+800 = 200(x^2+4x+4)#

The remaining quadratic factor is a perfect square trinomial:

#x^2+4x+4 = (x+2)^2#

Look at the pattern of the coefficients. It might remind you of this:

#144 = 12^2#

This is no coincidence. The coefficients are all small enough that no digits are carried during the multiplication of #12xx12#, so the digits #1,4,4# of #144# are the coefficients you get when you square #(x+2)#.

Putting it all together:

#200x^2+800x+800 = 200(x+2)^2#