How do you factor #45r^2-120rs+80s^2#?

1 Answer
Mar 30, 2016

Answer:

#45r^2-120rs+80s^2 = 5(3r-4s)^2#

Explanation:

First separate out the common scalar factor #5#:

#45r^2-120rs+80s^2=5(9r^2-24rs+16s^2)#

Next note that both #9r^2 = (3r)^2# and #16s^2 = (4s)^2# are perfect squares. So given the middle minus sign check to see if this is the square of #(3r-4s)#:

#(3r-4s)^2 = (3r)^2-2(3r)(4s)+(4s)^2 = 9r^2-24rs+16s^2#

So #9r^2-24rs+16s^2# is a perfect square trinomial.