How do you find the zeros of the function $f(x)=(x^2-x-12)/(x^2+2x-35)$ ?

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How do you find the zeros of the function $f(x)=(x^2-x-12)/(x^2+2x-35)$ ?

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Answer 1 · 2018-03-22T09:43:35

$x=-3" and "x=4$

Explanation:

$"to find the zeros equate the numerator to zero and solve"$

$"solve "x^2-x-12=0$

$"The factors of - 12 which sum to - 1 are - 4 and + 3"$

$rArr(x-4)(x+3)=0$

$"equate each factor to zero and solve for x"$

$x-4=0rArrx=4$

$x+3=0rArrx=-3$
graph{(x^2-x-12)/(x^2+2x-35 [-10, 10, -5, 5]}

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How do you find the zeros of the function $f(x)=(x^2-x-12)/(x^2+2x-35)$ ?

1 Answer

Explanation:

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Science

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How do you find the zeros of the function f(x)=(x^2-x-12)/(x^2+2x-35)f(x)=(x^2-x-12)/(x^2+2x-35)?

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Explanation:

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How do you find the zeros of the function $f(x)=(x^2-x-12)/(x^2+2x-35)$ ?