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

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

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Answer 1 · 2017-05-04T09:09:09

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

Explanation:

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

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

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

$\Rightarrow x^{3} + x^{2} - 6 x = 0 \to factor out x$ $rArrx^3+x^2-6x=0rarr" factor out x"$

$\Rightarrow x (x^{2} + x - 6) = 0$ $rArrx(x^2+x-6)=0$

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

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

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

1 Answer

Explanation:

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