Is it possible to factor y= x^5+2x^4+x^3+4x^2+5x+2 ? If so, what are the factors?

Dec 30, 2015

Yes, but it's a little complicated:

${x}^{5} + 2 {x}^{4} + {x}^{3} + 4 {x}^{2} + 5 x + 2$

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

$= \left(x + 2\right) \left({x}^{2} + \sqrt{\frac{\sqrt{17} - 1}{2}} x + \frac{\left(2 - \sqrt{\frac{\sqrt{17} - 1}{2}}\right) \left(\sqrt{17} + 1\right)}{8}\right)$

$\cdot \left({x}^{2} - \sqrt{\frac{\sqrt{17} - 1}{2}} x + \frac{\left(2 + \sqrt{\frac{\sqrt{17} - 1}{2}}\right) \left(\sqrt{17} + 1\right)}{8}\right)$

Explanation:

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

By the rational root theorem, the only possible rational roots of $f \left(x\right) = 0$ are $\pm 1$, $\pm 2$.

Trying these we find:

$f \left(- 2\right) = - 32 + 32 - 8 + 16 - 10 + 2 = 0$

So $\left(x + 2\right)$ is a factor:

${x}^{5} + 2 {x}^{4} + {x}^{3} + 4 {x}^{2} + 5 x + 2 = \left(x + 2\right) \left({x}^{4} + {x}^{2} + 2 x + 1\right)$

Next, since the remaining quartic factor has no ${x}^{3}$ term, it will factor into two quadratics with opposing middle coefficients:

${x}^{4} + {x}^{2} + 2 x + 1$

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

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

Equating coefficients and rearranging slightly, we get:

$b + c = {a}^{2} + 1$

$c - b = \frac{2}{a}$

$b c = 1$

So:

${\left({a}^{2} + 1\right)}^{2} = {\left(b + c\right)}^{2} = {\left(c - b\right)}^{2} + 4 b c = {\left(\frac{2}{a}\right)}^{2} + 4$

That is:

${\left({a}^{2}\right)}^{2} + 2 \left({a}^{2}\right) + 1 = \frac{4}{a} ^ 2 + 4$

Multiplying through by ${a}^{2}$ and rearranging a bit, that becomes:

${\left({a}^{2}\right)}^{3} + 2 {\left({a}^{2}\right)}^{2} - 3 \left({a}^{2}\right) - 4 = 0$

One of the solutions to this cubic in ${a}^{2}$ is ${a}^{2} = - 1$, but we would like to find a root that gives us a Real value for $a$, so let's factor this cubic and try again:

${\left({a}^{2}\right)}^{3} + 2 {\left({a}^{2}\right)}^{2} - 3 \left({a}^{2}\right) - 4 = \left({a}^{2} + 1\right) \left({\left({a}^{2}\right)}^{2} + {a}^{2} - 4\right)$

Hence two further roots:

${a}^{2} = \frac{- 1 \pm \sqrt{17}}{2}$

In particular, ${a}^{2} = \frac{\sqrt{17} - 1}{2}$, giving us:

$a = \pm \sqrt{\frac{\sqrt{17} - 1}{2}}$

We can use the positive square root, since the derivation is symmetric.

So, adding $b + c = {a}^{2} + 1$ and $c - b = \frac{2}{a}$

we find:

$2 c = {a}^{2} + 1 + \frac{2}{a}$

$= \frac{\sqrt{17} - 1}{2} + 1 + \frac{2}{\sqrt{\frac{\sqrt{17} - 1}{2}}}$

So:

$c = \frac{\sqrt{17} - 1}{4} + \frac{1}{2} + \frac{1}{\sqrt{\frac{\sqrt{17} - 1}{2}}}$

$= \frac{\sqrt{17} + 1}{4} + \frac{\sqrt{\frac{\sqrt{17} - 1}{2}}}{\frac{\sqrt{17} - 1}{2}}$

$= \frac{\sqrt{17} + 1}{4} + \frac{2 \sqrt{\frac{\sqrt{17} - 1}{2}}}{\sqrt{17} - 1}$

$= \frac{\sqrt{17} + 1}{4} + \frac{2 \sqrt{\frac{\sqrt{17} - 1}{2}} \left(\sqrt{17} + 1\right)}{16}$

$= \frac{\left(2 + \sqrt{\frac{\sqrt{17} - 1}{2}}\right) \left(\sqrt{17} + 1\right)}{8}$

Similarly:

$b = \frac{\left(2 - \sqrt{\frac{\sqrt{17} - 1}{2}}\right) \left(\sqrt{17} + 1\right)}{8}$

Putting it together:

${x}^{4} + {x}^{2} + 2 x + 1$

$= \left({x}^{2} + \sqrt{\frac{\sqrt{17} - 1}{2}} x + \frac{\left(2 - \sqrt{\frac{\sqrt{17} - 1}{2}}\right) \left(\sqrt{17} + 1\right)}{8}\right)$

$\cdot \left({x}^{2} - \sqrt{\frac{\sqrt{17} - 1}{2}} x + \frac{\left(2 + \sqrt{\frac{\sqrt{17} - 1}{2}}\right) \left(\sqrt{17} + 1\right)}{8}\right)$

Ouch!