How do you factor 10x^2-19xy+6y^2?

Jul 17, 2017

Because the sign are, $+ , - , +$, we know that the signs in the binomials are both minus:

$\left(a x - b y\right) \left(c x - \mathrm{dy}\right)$

We can write the following equations:

$a c = 10$
$b d = 6$
$a d + b c = 19$

Three equations and four unknown values is bad.

Let's start by trying 5 for a and 2 for c:

$\left(5 x - b y\right) \left(2 x - \mathrm{dy}\right)$

5d+2b=19
$b d = 6$

Two equations and two unknown values is good but does it work?

Is it $\left(2\right) \left(3\right) = 6$ or $\left(3\right) \left(2\right) = 6$?

$5 \left(3\right) + 2 \left(2\right) = 19$

It is $b = 2$ and $d = 3$:

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

Jul 17, 2017

(5x - 2y)(2x - 3y)

Explanation:

Consider x as a variable and y as a constant. Factor this quadratic trinomial by the new AC Method (Socratic, Google Search)
$f \left(x\right) = 10 {x}^{2} - 19 x y + 6 {y}^{2} =$ 10 (x + p)(x + q)
Converted trinomial:
$f ' \left(x\right) = {x}^{2} - 19 y x + 60 {y}^{2} =$ (x + p')(x + q').
Proceeding. Find p' and q', then, divide them by a = 10.
Find 2 numbers knowing the sum (b = - 19y) and product (60y^2).
They are: (-4y) and (-15y). (sum: -19y and product: 60y^2)
Back to f(x)--> $p = \frac{p '}{a} = - \frac{4 y}{10} = - \frac{2 y}{5}$, and $q = - \frac{15 y}{10} = - \frac{3 y}{2}$.
Factored form:
$f \left(x\right) = 10 \left(x - \frac{2 y}{5}\right) \left(x - \frac{3 y}{2}\right) = \left(5 x - 2 y\right) \left(2 x - 3 y\right)$