# How do you factor 4x^4-19x^3+16x^2-19x+12?

Oct 1, 2015

Use the rational root theorem, divide by the first factor found, then factor by grouping to find:

$4 {x}^{4} - 19 {x}^{3} + 16 {x}^{2} - 19 x + 12 = \left(4 x - 3\right) \left(x - 4\right) \left({x}^{2} + 1\right)$

#### Explanation:

Let $f \left(x\right) = 4 {x}^{4} - 19 {x}^{3} + 16 {x}^{2} - 19 x + 12$

By the rational root theorem, any rational roots of $f \left(x\right) = 0$ will be expressible in lowest terms as $\frac{p}{q}$, where $p , q \in \mathbb{Z}$, $q \ne 0$, $p$ a divisor of the constant term $12$ and $q$ a divisor of the coefficient $4$ of the leading term.

Also, since the signs of the terms are alternating, all the Real roots of $f \left(x\right) = 0$ are positive.

That means that the only possible rational roots are:

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

Let's try a few:

$f \left(\frac{1}{4}\right) = \frac{4}{256} - \frac{19}{64} + \frac{16}{16} - \frac{19}{4} + 12 = \frac{1 - 19 + 64 - 304 + 768}{64} = \frac{446}{64}$

$f \left(\frac{1}{2}\right) = \frac{4}{16} - \frac{19}{8} + \frac{16}{4} - \frac{19}{2} + 12 = \frac{2 - 19 + 32 - 76 + 96}{8} = \frac{35}{8}$

$f \left(\frac{3}{4}\right) = \frac{81}{64} - \frac{513}{64} + 9 - \frac{57}{4} + 12 = \frac{81 - 513 + 576 - 912 + 768}{64} = 0$

So $x = \frac{3}{4}$ is a root and $\left(4 x - 3\right)$ is a factor.

Divide by $\left(4 x - 3\right)$ to find:

$4 {x}^{4} - 19 {x}^{3} + 16 {x}^{2} - 19 x + 12 = \left(4 x - 3\right) \left({x}^{3} - 4 {x}^{2} + x - 4\right)$

The remaining cubic factor factors nicely by grouping:

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

$= {x}^{2} \left(x - 4\right) + 1 \left(x - 4\right) = \left({x}^{2} + 1\right) \left(x - 4\right)$

So putting this together, we have:

$4 {x}^{4} - 19 {x}^{3} + 16 {x}^{2} - 19 x + 12 = \left(4 x - 3\right) \left(x - 4\right) \left({x}^{2} + 1\right)$

The remaining quadratic factor ${x}^{2} + 1$ has no simpler linear factors with Real coefficients since ${x}^{2} + 1 \ge 1 > 0$ for all $x \in \mathbb{R}$.

If you really want to factor it further:

${x}^{2} + 1 = \left(x - i\right) \left(x + i\right)$

where $i$ is the imaginary unit.

Giving us:

$4 {x}^{4} - 19 {x}^{3} + 16 {x}^{2} - 19 x + 12 = \left(4 x - 3\right) \left(x - 4\right) \left(x - i\right) \left(x + i\right)$