# What are all the possible rational zeros for f(x)=2x^4-7x^3-2x^2-7x-4 and how do you find all zeros?

Oct 1, 2016

$f \left(x\right)$ has zeros $- \frac{1}{2}$, $4$, $\pm i$

#### Explanation:

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

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 4$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm 2 , \pm 4$

Before we embark on trying these, note that the coefficients $2 , - 2 , - 4$ almost cancel out if we reverse the sign of the middle one and the coefficients $- 7 , - 7$ would cancel out if the sign of one were opposite to the other. the only way that could heppen would be if ${x}^{2} = - 1$. So let's try $x = i$ first...

$f \left(i\right) = 2 {i}^{4} - 7 {i}^{3} - 2 {i}^{2} - 7 x - 4 = 2 + 7 i + 2 - 7 i - 4 = 0$

So $x = i$ is a zero and $\left(x - i\right)$ a factor.

Since the coefficients of $f \left(x\right)$ are all Real, $x = - i$ is also a zero and $\left(x + i\right)$ a factor.

Hence $\left(x - i\right) \left(x + i\right) = {x}^{2} + 1$ is a factor of $f \left(x\right)$...

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

We can factor the remaining quadratic using an AC method:

Find a pair of factors of $A C = 2 \cdot 4 = 8$ which differ by $B = 7$.

The pair $8 , 1$ works.

Use this pair to split the middle term and factor by grouping:

$2 {x}^{2} - 7 x - 4 = \left(2 {x}^{2} - 8 x\right) + \left(x - 4\right)$

$\textcolor{w h i t e}{2 {x}^{2} - 7 x - 4} = 2 x \left(x - 4\right) + 1 \left(x - 4\right)$

$\textcolor{w h i t e}{2 {x}^{2} - 7 x - 4} = \left(2 x + 1\right) \left(x - 4\right)$

Hence the other two zeros are:

$x = - \frac{1}{2}$ and $x = 4$