How do you find all the zeros of $f (x) = x^{3} + 6 x^{2} - 6 x - 36$ $f(x)=x^3 +6x^2 - 6x-36$ ?

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How do you find all the zeros of $f (x) = x^{3} + 6 x^{2} - 6 x - 36$ $f(x)=x^3 +6x^2 - 6x-36$ ?

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Answer 1 · 2016-03-05T03:57:51

Factor $f (x)$ $f(x)$ to see that the zeroes occur at $- 6$ $-6$ , $\sqrt{6}$ $sqrt(6)$ , and $- \sqrt{6}$ $-sqrt(6)$

Explanation:

Using the technique of factoring by grouping as well as the difference of squares formula , we can factor $f (x)$ $f(x)$ as

$x^{3} + 6 x^{2} - 6 x - 36 = x^{2} (x + 6) - 6 (x + 6)$ $x^3+6x^2-6x-36 = x^2(x+6)-6(x+6)$

$= (x^{2} - 6) (x + 6)$ $=(x^2-6)(x+6)$

$= (x + \sqrt{6}) (x - \sqrt{6}) (x + 6)$ $=(x+sqrt(6))(x-sqrt(6))(x+6)$

In its factored form, we can see that the zeros occur when any of the factors become $0$ $0$ , that is, when $x = - 6$ $x=-6$ or $x = \pm \sqrt{6}$ $x=+-sqrt(6)$ .

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How do you find all the zeros of $f (x) = x^{3} + 6 x^{2} - 6 x - 36$ $f(x)=x^3 +6x^2 - 6x-36$ ?

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How do you find all the zeros of f(x)=x3+6x2−6x−36f(x)=x^3 +6x^2 - 6x-36?

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Explanation:

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How do you find all the zeros of $f (x) = x^{3} + 6 x^{2} - 6 x - 36$ $f(x)=x^3 +6x^2 - 6x-36$ ?