# How do you find all the zeros of P(x) = x^4 + x^3 - 2x^2 + 4x – 24?

Jun 26, 2016

Zeros: $x = 2$, $x = - 3$, $x = \pm 2 i$

#### Explanation:

$P \left(x\right) = {x}^{4} + {x}^{3} - 2 {x}^{2} + 4 x - 24$

By the rational root theorem, any rational zeros of $P \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 24$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 8$, $\pm 12$, $\pm 24$

Trying each in turn, we find:

$P \left(1\right) = 1 + 1 - 2 + 4 - 24 = - 20$

$P \left(- 1\right) = 1 - 1 - 2 - 4 - 24 = - 30$

$P \left(2\right) = 16 + 8 - 8 + 8 - 24 = 0$

So $x = 2$ is a zero and $\left(x - 2\right)$ a factor:

${x}^{4} + {x}^{3} - 2 {x}^{2} + 4 x - 24 = \left(x - 2\right) \left({x}^{3} + 3 {x}^{2} + 4 x + 12\right)$

Note that the ratio between the first and second terms in the remaining cubic is the same as that between its third and fourth terms, so this cubic will factor by grouping:

${x}^{3} + 3 {x}^{2} + 4 x + 12$

$= \left({x}^{3} + 3 {x}^{2}\right) + \left(4 x + 12\right)$

$= {x}^{2} \left(x + 3\right) + 4 \left(x + 3\right)$

$= \left({x}^{2} + 4\right) \left(x + 3\right)$

Hence zero $x = - 3$

The remaining quadratic factor is always positive for Real values of $x$ so there are no more Real zeros, but it does have Complex zeros $x = \pm 2 i$, since:

${\left(2 i\right)}^{2} = 4 {i}^{2} = - 4$