# Determine the completely factored form of f(x) = 12x^3 - 44x^2 + 49x - 15?

Oct 21, 2016

$12 {x}^{3} - 44 {x}^{2} + 49 x - 15 = \left(2 x - 3\right) \left(3 x - 5\right) \left(2 x - 1\right)$

#### Explanation:

$f \left(x\right) = 12 {x}^{3} - 44 {x}^{2} + 49 x - 15$

By the rational roots theorem, any rational zero of $f \left(x\right)$ is expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 15$ and $q$ a divisor of the coefficient $12$ of the leading term.

In addition, note that the pattern of signs of coefficients of $f \left(x\right)$ is $+ - + -$ with $3$ sign changes, while those of $f \left(- x\right)$ are in the pattern $- - - -$ with $0$ sign changes. So by Descartes' Rule of Signs, $f \left(x\right)$ has $3$ or $1$ positive Real zero and no negative Real zeros.

Hence the only possible rational zeros are:

$\frac{1}{12} , \frac{1}{6} , \frac{1}{4} , \frac{1}{3} , \frac{5}{12} , \frac{1}{2} , \frac{3}{4} , \frac{5}{6} , 1 , \frac{5}{4} , \frac{3}{2} , \frac{5}{3} , \frac{5}{2} , 3 , 5 , 15$

We could simply try each of these in turn, but if you are allowed, we can speed up the process of finding the zeros as follows:

Look at the derivative and find out where it is $0$, indicating a local maximum or minimum...

$f ' \left(x\right) = 36 {x}^{2} - 88 x + 49$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\left(6 x - \frac{22}{3}\right)}^{2} - \frac{484}{9} + 49$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\left(6 x - \frac{22}{3}\right)}^{2} - {\left(\frac{\sqrt{43}}{3}\right)}^{2}$

$\textcolor{w h i t e}{f ' \left(x\right)} = \left(6 x - \frac{22}{3} - \frac{\sqrt{43}}{3}\right) \left(6 x - \frac{22}{3} + \frac{\sqrt{43}}{3}\right)$

Hence zero at $x = \frac{1}{6} \left(\frac{22}{3} \pm \frac{\sqrt{43}}{3}\right) = \frac{1}{18} \left(22 \pm \sqrt{43}\right)$

$\sqrt{43} \approx 6.5$

So the local maximum and minimum are approximately at:

$\frac{1}{18} \left(22 - 6.5\right) = \frac{31}{36} \approx \frac{5}{6}$ and $\frac{1}{18} \left(22 + 6.5\right) = \frac{19}{12} \approx \frac{3}{2}$

$f \left(\frac{5}{6}\right) = \frac{20}{9}$

$f \left(\frac{3}{2}\right) = 0$

Hmmm - a zero near where we expect the minimum. Let's look at the other rational possibility nearby...

$f \left(\frac{5}{3}\right) = 0$

So $x = \frac{3}{2}$ and $x = \frac{5}{3}$ are zeros with corresponding factors $\left(2 x - 3\right)$ and $\left(3 x - 5\right)$.

Looking at the coefficient of the leading term and constant term, the remaining factor (which must be rational) must be $\left(2 x - 1\right)$

So:

$12 {x}^{3} - 44 {x}^{2} + 49 x - 15 = \left(2 x - 3\right) \left(3 x - 5\right) \left(2 x - 1\right)$

graph{12x^3-44x^2+49x-15 [-0.448, 2.052, -1.26, 2.49]}