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

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Answer 1 · 2016-03-05T03:57:51

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

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

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

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

$=(x+sqrt(6))(x-sqrt(6))(x+6)$

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

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