# How do you factor 2x^4-9x^3+19x^2-15x?

Aug 19, 2016

$2 {x}^{4} - 9 {x}^{3} + 19 {x}^{2} - 15 x$

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

$= x \left(2 x - 3\right) \left(x - \frac{3}{2} - \frac{\sqrt{11}}{2} i\right) \left(x - \frac{3}{2} + \frac{\sqrt{11}}{2} i\right)$

#### Explanation:

$f \left(x\right) = 2 {x}^{4} - 9 {x}^{3} + 19 {x}^{2} - 15 x = x \left(2 {x}^{3} - 9 {x}^{2} + 19 x - 15\right)$

Let $g \left(x\right) = 2 {x}^{3} - 9 {x}^{2} + 19 x - 15$

By the rational roots theorem, any rational zeros of $g \left(x\right)$ (and therefore of $f \left(x\right)$) are 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 $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm \frac{5}{2} , \pm 3 , \pm 5 , \pm 15$

In addition note using Descartes' rule of signs, we find that this cubic has $1$ or $3$ positive Real zeros and no negative Real zeros.

So the only possible rational zeros are:

$\frac{1}{2} , 1 , \frac{3}{2} , \frac{5}{2} , 3 , 5 , 15$

Trying each in turn, we find:

$g \left(\frac{3}{2}\right) = 2 {\left(\frac{3}{2}\right)}^{3} - 9 {\left(\frac{3}{2}\right)}^{2} + 19 \left(\frac{3}{2}\right) - 15$

$= \frac{27}{4} - \frac{81}{4} + \frac{114}{4} - \frac{60}{4} = 0$

So $x = \frac{3}{2}$ is a zero and $\left(2 x - 3\right)$ a factor:

$2 {x}^{3} - 9 {x}^{2} + 19 x - 15 = \left(2 x - 3\right) \left({x}^{2} - 3 x + 5\right)$

The zeros of the remaining quadratic can be found using the quadratic formula:

$x = \frac{3 \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(5\right)}}{2 \cdot 1}$

$= \frac{3 \pm \sqrt{9 - 20}}{2}$

$= \frac{3}{2} \pm \frac{\sqrt{11}}{2} i$

Hence factors:

$\left(x - \frac{3}{2} - \frac{\sqrt{11}}{2} i\right) \left(x - \frac{3}{2} + \frac{\sqrt{11}}{2} i\right)$