How do you factor the trinomial #x^2 - 13xy + 36y^2#?

1 Answer
Dec 15, 2015

#x^2-13xy+36y^2 = (x-4y)(x-9y)#

Explanation:

You can solve the equation with respect to #x#, treating #y# as a normal number. So, we have that

#Delta = b^2-4ac = 169y^2 - 4*36y^2 = (169-144)y^2=25y^2#

#Delta# is a perfect squareand its root is #5y#. So, the solutions will be

#x_{1,2}=\frac{-b\pm\sqrt(Delta)}{2a} = \frac{13y\pm5y}{2}#

Which leads to #x_1 = (13+5)/2 y = 9y#, and #x_2 = (13-5)/2 y = 4y#

Once we know the two solutions, we can factor the quadratic expression with #(x-x_1)(x-x_2)#, so we have the factorization

#x^2-13xy+36y^2 = (x-4y)(x-9y)#