# How do you factor f^4-4f^2-9f+36?

Jan 14, 2018

${f}^{\textcolor{red}{3}} - 4 {f}^{2} - 9 f + 36 = \left(f - 3\right) \left(f + 3\right) \left(f - 4\right)$

#### Explanation:

Assuming a typo in the question, suppose we want to factor:

${f}^{\textcolor{red}{3}} - 4 {f}^{2} - 9 f + 36$

Note that the ratio between the first and second terms is the same as that between the third and fourth terms.

So this cubic will factor by grouping:

${f}^{3} - 4 {f}^{2} - 9 f + 36 = \left({f}^{3} - 4 {f}^{2}\right) - \left(9 f - 36\right)$

$\textcolor{w h i t e}{{f}^{3} - 4 {f}^{2} - 9 f + 36} = {f}^{2} \left(f - 4\right) - 9 \left(f - 4\right)$

$\textcolor{w h i t e}{{f}^{3} - 4 {f}^{2} - 9 f + 36} = \left({f}^{2} - 9\right) \left(f - 4\right)$

$\textcolor{w h i t e}{{f}^{3} - 4 {f}^{2} - 9 f + 36} = \left({f}^{2} - {3}^{2}\right) \left(f - 4\right)$

$\textcolor{w h i t e}{{f}^{3} - 4 {f}^{2} - 9 f + 36} = \left(f - 3\right) \left(f + 3\right) \left(f - 4\right)$

Jan 14, 2018

Here's one way you can try to do it...

#### Explanation:

If the question is correct in the given form, then this is how we can start to address it.

First note that there is no term in ${f}^{3}$ and hence there is a factorisation of the form:

${f}^{4} - 4 {f}^{2} - 9 f + 36 = \left({f}^{2} - a f + b\right) \left({f}^{2} + a f + c\right)$

$\textcolor{w h i t e}{{f}^{4} - 4 {f}^{2} - 9 f + 36} = {f}^{4} + \left(b + c - {a}^{2}\right) {f}^{2} + a \left(b - c\right) f + b c$

Equating coefficients we find:

$\left\{\begin{matrix}b + c = {a}^{2} - 4 \\ b - c = - \frac{9}{a} \\ b c = 36\end{matrix}\right.$

So:

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

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

$\textcolor{w h i t e}{{\left({a}^{2} - 4\right)}^{2}} = {\left(- \frac{9}{a}\right)}^{2} + 144$

That is:

${a}^{4} - 8 {a}^{2} + 16 = \frac{81}{a} ^ 2 + 144$

Multiplying both sides by ${a}^{2}$ and rearranging slightly, this becomes:

${\left({a}^{2}\right)}^{3} - 8 {\left({a}^{2}\right)}^{2} - 128 \left({a}^{2}\right) - 81 = 0$

I would like to simplify this to have no squared term. To avoid some fractions, multiply through by $27 = {3}^{3}$ first:

$0 = 27 {\left({a}^{2}\right)}^{3} - 216 {\left({a}^{2}\right)}^{2} - 3456 \left({a}^{2}\right) - 2187$

$\textcolor{w h i t e}{0} = {\left(3 {a}^{2} - 8\right)}^{3} - 1344 \left(3 a - 8\right) - 12427$

$\textcolor{w h i t e}{0} = {t}^{3} - 1344 t - 12427$

where $t = 3 {a}^{2} - 8$

This cubic in $t$ has $3$ real roots, which we can find with the help of a trigonometric substitution.

Let:

$t = 16 \sqrt{7} \cos \theta$

(The multiplier $16 \sqrt{7}$ is chosen so that the resulting expression contains $4 {\cos}^{3} \theta - 3 \cos \theta = \cos 3 \theta$)

Then:

$0 = {t}^{3} - 1344 t - 12427$

$\textcolor{w h i t e}{0} = 7168 \sqrt{7} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) - 12427$

$\textcolor{w h i t e}{0} = 7168 \sqrt{7} \cos 3 \theta - 12427$

So:

$\cos 3 \theta = \frac{12427}{7168 \sqrt{7}} = \frac{12427}{50176} \sqrt{7}$

Hence:

$3 \theta = \pm {\cos}^{- 1} \left(\frac{12427}{50176} \sqrt{7}\right) + 2 n \pi$

So:

$\theta = \pm \frac{1}{3} {\cos}^{- 1} \left(\frac{12427}{50176} \sqrt{7}\right) + \frac{2 n \pi}{3}$

This gives distinct solutions:

${t}_{n} = 16 \sqrt{7} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{12427}{50176} \sqrt{7}\right) + \frac{2 n \pi}{3}\right)$

for $n = 0 , 1 , 2$

The one we are interested in is ${t}_{0} \approx 40.6194$ since it is the only positive one.

Then we can choose:

$a = \sqrt{\frac{1}{3} \left(8 + 16 \sqrt{7} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{12427}{50176} \sqrt{7}\right)\right)\right)}$

Then:

$b = \frac{1}{2} \left({a}^{2} - 4 - \frac{9}{a}\right)$

$c = \frac{1}{2} \left({a}^{2} - 4 + \frac{9}{a}\right)$

Hence we have all of the coefficients of the quadratics to factor:

${f}^{2} - a f + b$

${f}^{2} + a f + c$

The details get very messy.