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

Mar 13, 2016

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

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

Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

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

$f \left(x\right) = {x}^{3} + 2 {x}^{2} - 9 x - 18$

$= \left({x}^{3} + 2 {x}^{2}\right) - \left(9 x + 18\right)$

$= {x}^{2} \left(x + 2\right) - 9 \left(x + 2\right)$

$= \left({x}^{2} - 9\right) \left(x + 2\right)$

$= \left({x}^{2} - {3}^{2}\right) \left(x + 2\right)$

$= \left(x - 3\right) \left(x + 3\right) \left(x + 2\right)$

Hence zeros:

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