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x^2 - 16 / 2x^2 - 5x - 12

 
klaensa1  Nov 10, 2017
edited by klaensa1  Nov 10, 2017

Best Answer 

 #1
avatar+302 
+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 \({x+4\over2x+3}\)wink

 
Mathhemathh  Nov 10, 2017
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1+0 Answers

 #1
avatar+302 
+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 \({x+4\over2x+3}\)wink

 
Mathhemathh  Nov 10, 2017

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