# How do you find all zeros with multiplicities of f(x)=2x^4+x^3-7x^2-3x+3?

Nov 5, 2017

The zeros are: $- 1 , \frac{1}{2} , \pm \sqrt{3}$, all with multiplicity $1$

#### Explanation:

Given:

$f \left(x\right) = 2 {x}^{4} + {x}^{3} - 7 {x}^{2} - 3 x + 3$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ can be expressed in the form $\frac{p}{q}$ for some integers $p , q$ with $p$ a divisor of the constant term $3$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 3$

We find:

$f \left(- 1\right) = 2 - 1 - 7 + 3 + 3 = 0$

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

$2 {x}^{4} + {x}^{3} - 7 {x}^{2} - 3 x + 3 = \left(x + 1\right) \left(2 {x}^{3} - {x}^{2} - 6 x + 3\right)$

Then note that the ratio of the first and second terms of the remaining cubic is the same as that between the third and fourth terms. So this cubic will factor by grouping:

$2 {x}^{3} - {x}^{2} - 6 x + 3 = \left(2 {x}^{3} - {x}^{2}\right) - \left(6 x - 3\right)$

$\textcolor{w h i t e}{2 {x}^{3} - {x}^{2} - 6 x + 3} = {x}^{2} \left(2 x - 1\right) - 3 \left(2 x - 1\right)$

$\textcolor{w h i t e}{2 {x}^{3} - {x}^{2} - 6 x + 3} = \left({x}^{2} - 3\right) \left(2 x - 1\right)$

$\textcolor{w h i t e}{2 {x}^{3} - {x}^{2} - 6 x + 3} = \left({x}^{2} - {\left(\sqrt{3}\right)}^{2}\right) \left(2 x - 1\right)$

$\textcolor{w h i t e}{2 {x}^{3} - {x}^{2} - 6 x + 3} = \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right) \left(2 x - 1\right)$

So the other zeros are $\pm \sqrt{3}$ and $\frac{1}{2}$