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

Question

Nathan Axel · Accepted Answer

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

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.

Ethan Adkins · Answer

undefined To factor the polynomial $x^2 - 8xy + 16y^2 - 3x + 12y + 2$, first, group the terms:

$(x^2 - 8xy + 16y^2) + (-3x + 12y + 2)$

Now, factor each group separately:

$x^2 - 8xy + 16y^2$ factors to $(x - 4y)^2$

$-3x + 12y + 2$ doesn't have a simple factorization.

So, the factored form of $x^2 - 8xy + 16y^2 - 3x + 12y + 2$ is $(x - 4y)^2 - 3x + 12y + 2$.