Question

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

Answer 1

`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)`