# What are the zeroes of f(x) = x^4 – x^3 + 7x^2 – 9x – 18?

Nov 28, 2015

$x = - 1$, $x = 3 i$, $x = - 3 i$ and $x = 2$

#### Explanation:

First notice that by reversing the signs of the coefficients of the terms with odd degree, the sum is zero. So $x = - 1$ is a zero:

$f \left(- 1\right) = 1 + 1 + 7 + 9 - 18 = 0$

and $\left(x + 1\right)$ is a factor:

${x}^{4} - {x}^{3} + 7 {x}^{2} - 9 x - 18$

$= \left(x + 1\right) \left({x}^{3} - 2 {x}^{2} + 9 x - 18\right)$

then factor by grouping...

$= \left(x + 1\right) \left(\left({x}^{3} - 2 {x}^{2}\right) + \left(9 x - 18\right)\right)$

$= \left(x + 1\right) \left({x}^{2} \left(x - 2\right) + 9 \left(x - 2\right)\right)$

$= \left(x + 1\right) \left({x}^{2} + 9\right) \left(x - 2\right)$

then take square root of $- 9$ to find:

$= \left(x + 1\right) \left(x - 3 i\right) \left(x + 3 i\right) \left(x - 2\right)$

So the zeros are $x = - 1$, $x = 3 i$, $x = - 3 i$ and $x = 2$