How do you find the zeros of the function $f(x)=(3x^2-18x+24)/(x-6)$ ?

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How do you find the zeros of the function $f(x)=(3x^2-18x+24)/(x-6)$ ?

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Answer 1 · 2017-06-05T07:34:20

$x=2,x=4$

Explanation:

$"zeros are the values of x that make f(x) equal zero"$

$rArr(3x^2-18x+24)/(x-6)=0$

$"the numerator therefore must equal zero"$

$"solve " 3x^2-18x+24=0" for zeros"$

$3(x^2-6x+8)=0larr" common factor of 3"$

$3(x-2)(x-4)=0$

$"equate each factor to zero"$

$x-2=0tox=2$

$x-4=0tox=4$

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How do you find the zeros of the function $f(x)=(3x^2-18x+24)/(x-6)$ ?

1 Answer

Explanation:

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How do you find the zeros of the function f(x)=(3x^2-18x+24)/(x-6)f(x)=(3x^2-18x+24)/(x-6)?

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Explanation:

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