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

Question

2299 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-15T12:12:56

$x^{2} - 13 x y + 36 y^{2} = (x - 4 y) (x - 9 y)$ $x^2-13xy+36y^2 = (x-4y)(x-9y)$

You can solve the equation with respect to $x$ $x$ , treating $y$ $y$ 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

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

How do you factor the trinomial x^2 - 13xy + 36y^2x2−13xy+36y2x^2 - 13xy + 36y^2?