How do you find all the zeros of $f(x) = x^3 + 2x^2 - 9x -18$ ?

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Answer 1 · 2016-03-13T00:31:02

Factor by grouping and using the difference of squares identity to find zeros:

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

The difference of squares identity can be written:

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

We use this with $a=x$ and $b=3$ below, but first factor by grouping:

$f(x) = x^3+2x^2-9x-18$

$=(x^3+2x^2)-(9x+18)$

$=x^2(x+2)-9(x+2)$

$=(x^2-9)(x+2)$

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

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

Hence zeros:

$x = 3$ , $x = -3$ and $x = -2$

How do you find all the zeros of f(x) = x^3 + 2x^2 - 9x -18f(x) = x^3 + 2x^2 - 9x -18?