# How do you find all the zeros of P(x)=x^4 + 1 x^3 - 3 x^2 - 1 x + 2?

Jun 2, 2016

#### Answer:

$x = 1$ with multiplicity $2$
$x = - 1$
$x = 2$

#### Explanation:

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

First note that the sum of the coefficients is $0$. That is:

$1 + 1 - 3 - 1 + 2 = 0$

So $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{4} + {x}^{3} - 3 {x}^{2} - x + 2 = \left(x - 1\right) \left({x}^{3} + 2 {x}^{2} - x - 2\right)$

Looking at the remaining cubic, notice that the ratio between the first and second terms is the same as that between the third and fourth terms. So it factors by grouping:

${x}^{3} + 2 {x}^{2} - x - 2$

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

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

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

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

Hence the remaining zeros are $1$, $- 1$, and $- 2$