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# what adds up to -5 but multiplies to make -12 ????

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1351
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+4

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

Nov 10, 2017
edited by klaensa1  Nov 10, 2017

#1
+349
+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 $${x+4\over2x+3}$$

Nov 10, 2017

#1
+349
+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