# How do you use synthetic division to find the zeroes of f(x)= x^4 -3x^3-9x^2-3x-10?

Jul 24, 2015

The zeroes of $f \left(x\right) = {x}^{4} - 3 {x}^{3} - 9 {x}^{2} - 3 x - 10$ are $\textcolor{red}{- 2 , 5 , - i , i}$.

#### Explanation:

According to the rational root theorem, the rational roots of $f \left(x\right) = 0$ must all be of the form $\frac{p}{q}$ with $p$ a divisor of $- 10$ and $q$ a divisor of $1$.

So the only possible rational roots are ±1,±2,±5,±10.

We have to test all eight possibilities.

Here are the only two that work. and So $- 2$ and $5$ are zeroes of the polynomial.

That means that $x + 2$ and $x - 5$ are factors, and

$\left(x + 2\right) \left(x - 5\right) = {x}^{2} - 3 x - 10$ is also a factor.

We can use synthetic division to find the other factor. The other factor is ${x}^{2} + 1$.

${x}^{2} + 1 = 0$
${x}^{2} = - 1$
x=±sqrt(-1) = ±i

$x = - i$ or $x = i$

So

$f \left(x\right) = {x}^{4} - 3 {x}^{3} - 9 {x}^{2} - 3 x - 10 = \left(x + 2\right) \left(x - 5\right) \left({x}^{2} + 1\right)$.

and

The roots of $f \left(x\right) = {x}^{4} - 3 {x}^{3} - 9 {x}^{2} - 3 x - 10$ are $- 2 , 5 , - i , i$.