#1**+2 **

I assume you have to simplify the expression by factoring. Okay...

\({x^2-16\over2x^2-5x-12}={(x+4)(x-4)\over(2x+3)(x-4)}={x+4\over2x+3}\).

To factor the denominator, you don't just find 2 numbers that add up and multiply to whatever if the coefficient of \(x^2\) is not 1. Since \(2x^2\) factors out to x and 2x, we have to find 2 numbers that, when one is added to **twice** the other, the result is -5 and when multiplied, the result is -12. The numbers that satisfy this are 3 and -4, so the denominator factors out to \((2x+3)(x-4)\) — the 2x is with the 3 becuase FOIL.

The numerator is easier, since it's just the product of the sum and difference of two numbers: \((x+4)(x-4)\). Cancelling out \((x-4)\), we get the final answer of \({x+4\over2x+3}\).

Mathhemathh
Nov 10, 2017

#1**+2 **

Best Answer

I assume you have to simplify the expression by factoring. Okay...

\({x^2-16\over2x^2-5x-12}={(x+4)(x-4)\over(2x+3)(x-4)}={x+4\over2x+3}\).

To factor the denominator, you don't just find 2 numbers that add up and multiply to whatever if the coefficient of \(x^2\) is not 1. Since \(2x^2\) factors out to x and 2x, we have to find 2 numbers that, when one is added to **twice** the other, the result is -5 and when multiplied, the result is -12. The numbers that satisfy this are 3 and -4, so the denominator factors out to \((2x+3)(x-4)\) — the 2x is with the 3 becuase FOIL.

The numerator is easier, since it's just the product of the sum and difference of two numbers: \((x+4)(x-4)\). Cancelling out \((x-4)\), we get the final answer of \({x+4\over2x+3}\).

Mathhemathh
Nov 10, 2017