How do you factor 30x^5y-85x^3y^2+25xy^3?

May 13, 2016

$5 x y \left(y - 2 {x}^{2}\right) \left(y - 3 {x}^{2}\right)$

Explanation:

$F \left(x , y\right) = 5 x y f \left(x , y\right) = 5 x y \left(5 {y}^{2} - 17 {x}^{2} y + 6 {x}^{4}\right) .$
Factor f(x, y) by considering y as variable and x as constant.
Use the new AC Method to factor trinomials (Socratic Search).
$f \left(x , y\right) = 5 {y}^{2} - 17 {x}^{2} y + 6 {x}^{4} =$ 5(y + p)(y + q)
Converted trinomial: $f ' \left(x , y\right) = {y}^{2} - 17 {x}^{2} y + 30 {x}^{4} =$ (y + p')(y + q')
p' and q' have same sign because ac > 0.
Factor pairs of $\left(a c = 30 {x}^{4}\right)$ --> $\left(- 2 {x}^{2} , - 15 {x}^{2}\right)$. This sum is $- 17 {x}^{2} = b$. Therefor, $p ' = - 2 {x}^{2}$ and $q ' = - 15 {x}^{2}$.
Back to original f(x, y) --> $p = \frac{p '}{a} = - \frac{2 {x}^{2}}{5}$ and
$q = \frac{q '}{a} = - \frac{15 {x}^{2}}{5} = - 3 {x}^{2}$
Factored form of $f \left(x\right) = 5 \left(y - \frac{2 {x}^{2}}{5}\right) \left(y - 3 {x}^{2}\right) = \left(5 y - 2 {x}^{2}\right) \left(y - 3 {x}^{2}\right)$
$F \left(x\right) = 5 x y . f \left(x\right) = 5 x y \left(y - 2 {x}^{2}\right) \left(y - 3 {x}^{2}\right)$