How do you factor 9x^2 - 18xy + 5y^2?

1 Answer
May 16, 2015

9x^2-18xy+5y^2 = (3x-5y)(3x-y)

To find this, suppose (9x^2-18xy+5y^2)=(ax+by)(cx+dy)

Then from multiplying out the right hand side and comparing coefficients we find:

9 = ac

-18 = ad+bc

5=bd

We might as well choose a and c to both be positive.

bd=5 tells us that either both b and d are positive or b and d are both negative.

Since we have chosen a and c to be positive, we can deduce from the -18 = ad+bc equation that b and d are both negative.

So bd = 5 only factors as -5 xx -1 or -1 xx -5.

As everything so far is symmetric, let's choose b=-5 and d=-1

ac=9 will only factor as 1xx9, 3xx3 or 9xx1.

Trying each of these possibilities, we find that -18=ad+bc is satisfied when a=3 and c=3.