# How do you find all rational zeroes of the function using synthetic division f(x)=3x^3+12x^2+3x-18?

Apr 16, 2017

The "possible" rational zeros are $\pm 1 , \pm 2 , \pm 3 , \pm 6$

The actual zeros are: $1$, $- 2$ and $- 3$

#### Explanation:

Given:

$f \left(x\right) = 3 {x}^{3} + 12 {x}^{2} + 3 x - 18$

First note that all of the coefficients are divisible by $3$, so separate that out as a factor...

$3 {x}^{3} + 12 {x}^{2} + 3 x - 18 = 3 \left({x}^{3} + 4 {x}^{2} + x - 6\right)$

By the rational roots theorem, any rational zeros of ${x}^{3} + 4 {x}^{2} + x - 6$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 6$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 6$

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

$1 + 4 + 1 - 6 = 0$

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

We can use synthetic division to find:

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

It looks something like this:

$\underline{1 \textcolor{w h i t e}{0} |} \textcolor{w h i t e}{00} 1 \textcolor{w h i t e}{00} 4 \textcolor{w h i t e}{00} 1 \textcolor{w h i t e}{00} - 6$
$\textcolor{w h i t e}{00 |} \underline{\textcolor{w h i t e}{00000} 1 \textcolor{w h i t e}{00} 5 \textcolor{w h i t e}{00 -} 6}$
$\textcolor{w h i t e}{00 | 0} 1 \textcolor{w h i t e}{00} 5 \textcolor{w h i t e}{00} 6 \textcolor{w h i t e}{00 -} \textcolor{b l u e}{0}$

where the final $\textcolor{b l u e}{0}$ shows us that the remainder is $0$ as expected.

To factor the remaining quadratic, note that $5 = 2 + 3$ and $6 = 2 \cdot 3$, so:

${x}^{2} + 5 x + 6 = \left(x + 2\right) \left(x + 3\right)$

So:

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

with zeros $1$, $- 2$ and $- 3$.