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

2 Answers
Jul 17, 2017

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)#

Jul 17, 2017

Answer:

(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)#