# What are all the possible rational zeros for f(x)=x^4-3x^3-6x^2+6x+8 and how do you find all zeros?

Nov 14, 2016

The "possible" rational zeros are: $\pm 1$, $\pm 2$, $\pm 4$, $\pm 8$

The actual zeros are: $- 1$, $4$, $\pm \sqrt{2}$

#### Explanation:

$f \left(x\right) = {x}^{4} - 3 {x}^{3} - 6 {x}^{2} + 6 x + 8$

By the rational root 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 $8$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 8$

We find:

$f \left(- 1\right) = 1 + 3 - 6 - 6 + 8 = 0$

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

${x}^{4} - 3 {x}^{3} - 6 {x}^{2} + 6 x + 8 = \left(x + 1\right) \left({x}^{3} - 4 {x}^{2} - 2 x + 8\right)$

Notice that the ratio between 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:

${x}^{3} - 4 {x}^{2} - 2 x + 8 = \left({x}^{3} - 4 {x}^{2}\right) - \left(2 x - 8\right)$

$\textcolor{w h i t e}{{x}^{3} - 4 {x}^{2} - 2 x + 8} = {x}^{2} \left(x - 4\right) - 2 \left(x - 4\right)$

$\textcolor{w h i t e}{{x}^{3} - 4 {x}^{2} - 2 x + 8} = \left({x}^{2} - 2\right) \left(x - 4\right)$

Finally we can factor the quadratic as a difference of squares:

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

So the zeros are: $- 1$, $4$, $\pm \sqrt{2}$