How do you factor # x^2 - 8xy + 16y^2 - 3x + 12y +2#?

1 Answer
May 11, 2016

Answer:

#x^2-8xy+16y^2-3x+12y+2=(x-4y-1)(x-4y-2)#

Explanation:

This is a disguised version of:

#t^2-3t+2 = (t-1)(t-2)#

with #t=x-4y# as follows:

#x^2-8xy+16y^2-3x+12y+2#

#=(x-4y)^2-3(x-4y)+2#

#=((x-4y)-1)((x-4y)-2)#

#=(x-4y-1)(x-4y-2)#

A Little Slower

This polynomial is a mixture of terms of degree #2#, #1# and #0#.

So if it factors, then it has two factors each containing a mixture of terms of degree #1# and #0#.

If we removed the terms of degree #0# from both of the factors, then the product of the simplified factors would be exactly the terms of degree #2#.

So to find the terms of degree #1# in each factor we just need to look at the terms of degree #2# in the original polynomial, namely:

#x^2-8xy+16y^2#

Note that #x^2# and #16y^2 = (4y)^2# are both perfect squares, so we might hope and indeed find that this is a perfect square trinomial:

#x^2-8xy+16y^2 = (x-4y)^2#

Next note that the terms of degree #1# in the original polynomial are a simple scalar multiple of the same #(x-4y)#, namely #-3(x-4y)#

Hence we find:

#x^2-8xy+16y^2-3x+12y+2 = (x-4y)^2-3(x-4y)+2#

Then substitute #t = (x-4y)#, to get #t^2-3t+2#, which factorises as #(t-1)(t-2)#, etc.