How do you find the zeros of the function $f(x)=(x^3+x^2-6x)/(x-1)$ ?

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How do you find the zeros of the function $f(x)=(x^3+x^2-6x)/(x-1)$ ?

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Answer 1 · 2017-05-04T09:09:09

$x=-3,x=0,x=2$

Explanation:

$"the zeros are the values of x that make f(x)=0"$

$"The denominator of "f(x)!=0$
$"as this would make f(x) undefined"$

$"equate the numerator to zero and solve"$

$rArrx^3+x^2-6x=0rarr" factor out x"$

$rArrx(x^2+x-6)=0$

$rArrx(x+3)(x-2)=0$

$rArrx=0,x=-3,x=2larr" are the zeros"$

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How do you find the zeros of the function $f(x)=(x^3+x^2-6x)/(x-1)$ ?

1 Answer

Explanation:

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How do you find the zeros of the function f(x)=(x^3+x^2-6x)/(x-1)f(x)=(x^3+x^2-6x)/(x-1)?

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How do you find the zeros of the function $f(x)=(x^3+x^2-6x)/(x-1)$ ?