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

Question

Julia Adams · Accepted Answer

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)#

Savannah Anderson · Answer

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

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)#

Nathan Axel · Answer

undefined To factor the expression $10x^2 - 19xy + 6y^2$, we first look for two numbers that multiply to $10 \times 6 = 60$ and add up to $-19$. These numbers are $-15$ and $-4$. Then, we rewrite the middle term using these numbers, which gives us $10x^2 - 15xy - 4xy + 6y^2$. We group the terms and factor by grouping, yielding $5x(2x - 3y) - 2y(2x - 3y)$. Finally, we factor out the common factor $(2x - 3y)$ to get $(5x - 2y)(2x - 3y)$.