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

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

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Answer 1 · 2018-03-22T09:43:35

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

Explanation:

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

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

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

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

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

$x - 4 = 0 \Rightarrow x = 4$ $x-4=0rArrx=4$

$x + 3 = 0 \Rightarrow x = - 3$ $x+3=0rArrx=-3$
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 $f (x) = \frac{x^{2} - x - 12}{x^{2} + 2 x - 35}$ $f(x)=(x^2-x-12)/(x^2+2x-35)$ ?

1 Answer

Explanation:

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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)=x2−x−12x2+2x−35f(x)=(x^2-x-12)/(x^2+2x-35)?

1 Answer

Explanation:

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