# How do you use synthetic division to find the factors of f(x)= x^4 -x^3 -19x^2+49x-30?

Aug 30, 2015

If you can spot a factor of $f \left(x\right)$ then you can use synthetic division to divide by that factor to give a simpler polynomial to factor.

Hence we can find:

$f \left(x\right) = \left(x - 1\right) \left(x + 5\right) \left(x - 2\right) \left(x - 3\right)$

#### Explanation:

First note that $f \left(1\right) = 1 - 1 - 19 + 49 - 30 = 0$, so $\left(x - 1\right)$ is a factor of $f \left(x\right)$

Use synthetic division to divide $f \left(x\right)$ by $\left(x - 1\right)$:

So $f \left(x\right) = \left(x - 1\right) \left({x}^{3} - 19 x + 30\right)$

Notice that ${\left(- 5\right)}^{3} - 19 \cdot \left(- 5\right) + 30 = - 125 + 95 + 30 = 0$, so $\left(x + 5\right)$ is also a factor of $f \left(x\right)$.

Divide ${x}^{3} - 19 x + 30$ by $\left(x + 5\right)$ using synthetic division - not forgetting to specify the coefficient $0$ of the ${x}^{2}$ term:

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

By this stage you can probably spot that ${x}^{2} - 5 + 6 = \left(x - 2\right) \left(x - 3\right)$ to complete our factorisation:

$f \left(x\right) = \left(x - 1\right) \left(x + 5\right) \left(x - 2\right) \left(x - 3\right)$