How do you simplify #\frac { 2x ^ { 2} + 5x y - 12y ^ { 2} } { 2x ^ { 2} + 9x y + 4y ^ { 2} }#?

2 Answers
Jan 24, 2018

This gives #(2x - 3y)/(2x + y)#

Explanation:

Learn to factor trinomials. When you want to factor something like #2x^2 -7x + 5#, you want to find two numbers that multiply to the coefficients of the highest and lowest degree term and add to the middle degree term.

Two numbers that multiply to #+10# and add to #-7# are #-5# and #-2#. Therefore, #2x^2 - 7x + 5 = 2x^2 - 2x - 5x + 5 = 2x(x - 1) - 5(x - 1) = (2x- 5)(x - 1)#

We will apply this to our problem now. Two numbers that multiply to #-24# and add to #+5# would be #+8# and #-3#. Thus:

#2x^2 +5xy -12y^2 =2x^2 + 8xy - 3xy - 12y^2= 2x(x + 4y) - 3y(x + 4y) = (2x - 3y)(x + 4y)#

Now to the denominator. Two numbers that multiply to #8# and add to #9# would be #+8# and #+1#.

#2x^2 + 9xy +4y^2 =2x^2 +8xy +xy + 4y^2 =2x(x + 4y) + y(x + 4y) = (2x + y)(x + 4y)#

Rewriting in fraction form:

#((2x - 3y)(x + 4y))/((2x + y)(x + 4y))#

#= (2x- 3y)/(2x +y)#

Hopefully this helps!

Jan 24, 2018

#(​2x−3y)/(2x-y)​#

Explanation:

first factor the top equation to: #2x^2-3xy+8xy-12y^2# then group then and and factor out to get: #(x+4y)(2x-3y)# then do the same for the second equation and you'll get: #(2x+y)(x+4y)# notice that the two equation both have #(x+4y)# in common and so they cancel each other out. so you'll let with your answer