# How do you factor given that f(4)=0 and f(x)=x^3-14x^2+47x-18?

Aug 2, 2016

$f \left(x\right) = \left(x - 9\right) \left(x - \frac{5}{2} - \frac{\sqrt{17}}{2}\right) \left(x - \frac{5}{2} + \frac{\sqrt{17}}{2}\right)$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We use this later with $a = \left(x - \frac{5}{2}\right)$ and $b = \frac{\sqrt{17}}{2}$.

By the rational root theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 18$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 6 , \pm 9 , \pm 18$

In particular $f \left(4\right) \ne 0$ since $4$ is not a factor of $18$.

Trying each of these in turn, we (eventually) find:

$f \left(9\right) = 729 - 1134 + 423 - 18 = 0$

So $\left(x - 9\right)$ is a factor of $f \left(x\right)$:

${x}^{3} - 14 {x}^{2} + 47 x - 18$

$= \left(x - 9\right) \left({x}^{2} - 5 x + 2\right)$

$= \left(x - 9\right) \left({\left(x - \frac{5}{2}\right)}^{2} - \frac{25}{4} + 2\right)$

$= \left(x - 9\right) \left({\left(x - \frac{5}{2}\right)}^{2} - {\left(\frac{\sqrt{17}}{2}\right)}^{2}\right)$

$= \left(x - 9\right) \left(x - \frac{5}{2} - \frac{\sqrt{17}}{2}\right) \left(x - \frac{5}{2} + \frac{\sqrt{17}}{2}\right)$