# How do you factor y^4+8y^3+12?

Feb 20, 2017

If the question is correct as stands then the answer is complicated, but here's a sketch of how you can do it...

#### Explanation:

${y}^{4} + 8 {y}^{3} + 12$

Let $t = y + 2$

Then:

${y}^{4} + 8 {y}^{3} + 12 = {\left(y + 2\right)}^{4} - 24 {\left(y + 2\right)}^{2} + 64 \left(y + 2\right) - 36$

$\textcolor{w h i t e}{{y}^{4} + 8 {y}^{3} + 12} = {t}^{4} - 24 {t}^{2} + 64 t - 36$

$\textcolor{w h i t e}{{y}^{4} + 8 {y}^{3} + 12} = \left({t}^{2} - a t + b\right) \left({t}^{2} + a t + c\right)$

$\textcolor{w h i t e}{{y}^{4} + 8 {y}^{3} + 12} = {t}^{4} + \left(b + c - {a}^{2}\right) {t}^{2} + a \left(b - c\right) t + b c$

Equating coefficients:

$\left\{\begin{matrix}b + c = {a}^{2} - 24 \\ b - c = \frac{64}{a} \\ b c = - 36\end{matrix}\right.$

Hence:

${a}^{4} - 48 {a}^{2} + 576 = {\left({a}^{2} - 24\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = \frac{4096}{a} ^ 2 - 144$

Hence:

$0 = {\left({a}^{2}\right)}^{3} - 48 {\left({a}^{2}\right)}^{2} + 720 \left({a}^{2}\right) - 4096$

$\textcolor{w h i t e}{0} = {\left({a}^{2} - 16\right)}^{3} - 48 \left({a}^{2} - 16\right) - 768$

Using Cardano's method, let ${a}^{2} - 16 = u + v$

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 16\right) \left(u + v\right) - 768 = 0$

Add the constraint $v = \frac{16}{u}$ to eliminate the term in $\left(u + v\right)$ and get:

${\left({u}^{3}\right)}^{2} - 768 \left({u}^{3}\right) + 4096 = 0$

${u}^{3} = \frac{768 \pm \sqrt{{768}^{2} - 4 \left(1\right) \left(4096\right)}}{2}$

$\textcolor{w h i t e}{{u}^{3}} = 384 \pm 64 \sqrt{35}$

$\textcolor{w h i t e}{{u}^{3}} = {4}^{3} \left(6 \pm \sqrt{35}\right)$

Hence:

${a}^{2} = 16 + 4 \sqrt{6 + \sqrt{35}} + 4 \sqrt{6 - \sqrt{35}}$

We can use the positive square root:

$a = 2 \sqrt{4 + \sqrt{6 + \sqrt{35}} + \sqrt{6 - \sqrt{35}}}$

Then:

$b = \frac{1}{2} \left({a}^{2} - 24 + \frac{64}{a}\right)$

$c = \frac{1}{2} \left({a}^{2} - 24 - \frac{64}{a}\right)$

Leaving us with two quadratics to solve:

${t}^{2} - a t + b = 0$

${t}^{2} + a t + c = 0$

yielding $4$ roots: ${t}_{1} , {t}_{2} , {t}_{3} , {t}_{4}$.

Then the zeros of our original quartic are given by:

${y}_{n} = {t}_{n} - 2$

Hence we find a factorisation:

${y}^{4} + 8 {y}^{3} + 12 = \left(y - {y}_{1}\right) \left(y - {y}_{2}\right) \left(y - {y}_{3}\right) \left(y - {y}_{4}\right)$

where ${y}_{1} , {y}_{2} , {y}_{3} , {y}_{4}$ are the four zeros.