# What are all the possible rational zeros for f(x)=3x^4-10x^3-24x^2-6x+5 and how do you find all zeros?

Sep 28, 2016

Use the rational root theorem and the remainder theorem to determine a zero.

The possible factors in $y = a {x}^{n} + b {x}^{n - 1} + \ldots + p$ are given by $\text{factors of p"/"factors of a}$.

In our example, these would be $\pm 1 , \pm \frac{5}{3} , \pm \frac{5}{1} \pm \frac{1}{3} \to \pm 1 , \pm \frac{5}{3} \pm 5 , \pm \frac{1}{3}$.

We now use the remainder theorem to determine the remainder after substituting these values into the function. If the remainder equals 0, we have a factor.

$f \left(- 1\right) = 3 {\left(- 1\right)}^{4} - 10 {\left(- 1\right)}^{3} - 24 {\left(- 1\right)}^{2} - 6 \left(- 1\right) + 5$

$f \left(- 1\right) = 3 \left(1\right) - 10 \left(- 1\right) - 24 \left(1\right) + 6 + 5$

$f \left(- 1\right) = 3 + 10 - 24 + 6 + 5$

$f \left(- 1\right) = 0$

Hence, $- 1$ is a factor. Next, we need to use synthetic division to divide $3 {x}^{4} - 10 {x}^{3} - 24 {x}^{2} - 6 x + 5$ by $x + 1$

$- 1 \text{_|"3" "-10" "-24" "-6" 5}$
$\text{ "-3" "13" "11" "-5}$
$\text{-----------------------------------------------------------------------------}$
$\text{ "3" "-13" "-11" "5" 0 }$

Hence, the quotient is $3 {x}^{3} - 13 {x}^{2} - 11 x + 5$.

We repeat the process to find that $- 1$ is once again a factor. We divide synthetically again, to obtain the result of $3 {x}^{2} - 16 x + 5$.

We can factor this as:

$3 {x}^{2} - 16 x + 5 = 3 {x}^{2} - 15 x - x + 5 = 3 x \left(x - 5\right) - 1 \left(x - 5\right) = \left(3 x - 1\right) \left(x - 5\right)$

Setting all of these factors to $0$, we have that this function has a root of $x = - 1$ of multiplicity $2$, and roots of $\frac{1}{3}$ and $5$ of multiplicity $1$.

Here is the graph of the function. Hopefully this helps!