How do you factor 2x^4-19x^3+57x^2-64x+20?

Jan 30, 2016

Use the rational root theorem to help find the first couple of factors and separate them out to find:

$2 {x}^{4} - 19 {x}^{3} + 57 {x}^{2} - 64 x + 20 = \left(2 x - 1\right) \left(x - 2\right) \left(x - 2\right) \left(x - 5\right)$

Explanation:

Let $f \left(x\right) = 2 {x}^{4} - 19 {x}^{3} + 57 {x}^{2} - 64 x + 20$

By the rational root theorem any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p$ and $q$ with $p$ a divisor of the constant term $20$ and $q$ a divisor of the coefficient $2$ of the leading term.

So the only possible rational roots are:

$\pm \frac{1}{2}$, $\pm 1$, $\pm 2$, $\pm \frac{5}{2}$, $\pm 4$, $\pm 5$, $\pm 10$, $\pm 20$

Trying these in turn we find:

$f \left(\frac{1}{2}\right) = \frac{1}{8} - \frac{19}{8} + \frac{57}{4} - 32 + 20 = \frac{1 - 19 + 114 - 256 + 160}{8} = 0$

So $x = \frac{1}{2}$ is a zero and $\left(2 x - 1\right)$ is a factor:

$2 {x}^{4} - 19 {x}^{3} + 57 {x}^{2} - 64 x + 20 = \left(2 x - 1\right) \left({x}^{3} - 9 {x}^{2} + 24 x - 20\right)$

Having 'used up' the factor of $2$ on the leading coefficient, the remaining possible rational roots are:

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 5$, $\pm 10$, $\pm 20$

Let $g \left(x\right) = {x}^{3} - 9 {x}^{2} + 24 x - 20$

We find:

$g \left(1\right) = 1 - 9 + 24 - 20 = - 4$

$g \left(- 1\right) = - 1 - 9 - 24 - 20 = - 54$

$g \left(2\right) = 8 - 36 + 48 - 20 = 0$

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

$\left(2 x - 1\right) \left({x}^{3} - 9 {x}^{2} + 24 x - 20\right) = \left(2 x - 1\right) \left(x - 2\right) \left({x}^{2} - 7 x + 10\right)$

The remaining quadratic can be factored by noticing that $2 + 5 = 7$ and $2 \times 5 = 10$, hence:

$\left(2 x - 1\right) \left(x - 2\right) \left({x}^{2} - 7 x + 10\right) = \left(2 x - 1\right) \left(x - 2\right) \left(x - 2\right) \left(x - 5\right)$

graph{2x^4-19x^3+57x^2-64x+20 [-7.42, 12.58, -5.8, 4.2]}