# How do you factor 2x^4 + 5x^3 - 2x^2 + 5x + 3?

Apr 16, 2016

See explanation...

#### Explanation:

It is possible, but horribly messy to solve this algebraically.

Here's the beginning of a solution, to show you the direction and how complicated it gets. Note that I may have made some arithmetic errors, but the method illustrated is sound:

$f \left(x\right) = 2 {x}^{4} + 5 {x}^{3} - 2 {x}^{2} + 5 x + 3$

To factor this, first multiply by ${2}^{11} = 2048$ to get:

$2048 f \left(x\right) = 4096 {x}^{4} + 10240 {x}^{3} - 4096 {x}^{2} + 10240 x + 6144$

$= {\left(8 x + 5\right)}^{4} - 214 {\left(8 x + 5\right)}^{2} + 994 \left(8 x + 5\right) + 5557$

Let $t = 8 x + 5$

Then we want to find the zeros of:

${t}^{4} - 214 {t}^{2} + 994 t + 5557$

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

$= {t}^{4} + \left(b + c - {a}^{2}\right) {t}^{2} + \left(c - b\right) a t + b c$

Equating coefficients we find:

$\left\{\begin{matrix}b + c = {a}^{2} - 214 \\ c - b = \frac{994}{a} \\ b c = 5557\end{matrix}\right.$

Then:

${\left({a}^{2} - 214\right)}^{2} = {\left(b + c\right)}^{2} = {\left(c - b\right)}^{2} + 4 b c = {\left(\frac{994}{a}\right)}^{2} + \left(4 \cdot 5557\right)$

Hence:

${\left({a}^{2}\right)}^{3} - 428 {\left({a}^{2}\right)}^{2} + 23568 \left({a}^{2}\right) - 988036 = 0$

Multiply through by $27$ to get:

$27 {\left({a}^{2}\right)}^{3} - 11556 {\left({a}^{2}\right)}^{2} + 636336 \left({a}^{2}\right) - 26676972 = 0$

Which becomes:

${\left(3 {a}^{2} - 428\right)}^{3} - 337440 \left(3 {a}^{2} - 428\right) - 92698540 = 0$

Let $s = 3 {a}^{2} - 428$, so:

${s}^{3} - 337440 s - 92698540 = 0$

Let $s = u + v$:

${u}^{3} + {v}^{3} + 3 \left(u v - 112480\right) \left(u + v\right) - 92698540 = 0$

Add the constraint $v = \frac{112480}{u}$ to derive:

${\left({u}^{3}\right)}^{2} - 92698540 \left({u}^{3}\right) + 1423068884992000 = 0$

Using the quadratic formula and the symmetry in $u$ and $v$ we can find the Real root:

$s = \sqrt{46349270 + 30 \sqrt{805762160601}} + \sqrt{46349270 - 30 \sqrt{805762160601}}$

Hence:

${a}^{2} = \frac{s + 428}{3}$

$= \frac{1}{3} \left(428 + \sqrt{46349270 + 30 \sqrt{805762160601}} + \sqrt{46349270 - 30 \sqrt{805762160601}}\right)$

We can then feed this value for $a$ back into our simultaneous equations to find $b$ and $c$, hence some quadratics to solve.