Question

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

Answer 1

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

`(ax-by)(cx-dy)`

We can write the following equations:

`ac = 10`
`bd = 6`
`ad+bc = 19`

Three equations and four unknown values is bad.

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

`(5x-by)(2x-dy)`

5d+2b=19
`bd = 6`

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

Is it `(2)(3) = 6` or `(3)(2)=6`?

`5(3)+2(2)=19`

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

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

Answer 2

(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(x) = 10x^2 - 19xy + 6y^2 =` 10 (x + p)(x + q)
Converted trinomial:
`f'(x) = x^2 - 19yx + 60y^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 = (p')/a = - (4y)/10 = -(2y)/5`, and `q = - (15y)/10 = - (3y)/2`.
Factored form:
`f(x) = 10(x - (2y)/5)(x - (3y)/2) = (5x - 2y)(2x - 3y)`