Question

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

Answer 1

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!

Answer 2

`(​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