How do you find the zeros of $x^3+x^2-4x-4$ ?

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Answer 1 · 2016-08-19T23:11:52

This cubic has zeros: $2, -2, -1$

This cubic factors by grouping then using the difference of squares identity:

$a^2-b^2=(a-b)(a+b)$

with $a=x$ and $b=2$ as follows:

$x^3+x^2-4x-4$

$=(x^3+x^2)-(4x+4)$

$=x^2(x+1)-4(x+1)$

$=(x^2-4)(x+1)$

$=(x^2-2^2)(x+1)$

$=(x-2)(x+2)(x+1)$

Hence zeros:

$2, -2, -1$

How do you find the zeros of x^3+x^2-4x-4x^3+x^2-4x-4?