Question

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

Answer 1

`f^color(red)(3)-4f^2-9f+36 = (f-3)(f+3)(f-4)`

Explanation:

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

`f^color(red)(3)-4f^2-9f+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^3-4f^2-9f+36 = (f^3-4f^2)-(9f-36)`

`color(white)(f^3-4f^2-9f+36) = f^2(f-4)-9(f-4)`

`color(white)(f^3-4f^2-9f+36) = (f^2-9)(f-4)`

`color(white)(f^3-4f^2-9f+36) = (f^2-3^2)(f-4)`

`color(white)(f^3-4f^2-9f+36) = (f-3)(f+3)(f-4)`

Answer 2

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^4-4f^2-9f+36 = (f^2-af+b)(f^2+af+c)`

`color(white)(f^4-4f^2-9f+36) = f^4+(b+c-a^2)f^2+a(b-c)f+bc`

Equating coefficients we find:

`{ (b+c = a^2-4), (b-c = -9/a), (bc=36) :}`

So:

`(a^2-4)^2 = (b+c)^2`

`color(white)((a^2-4)^2) = (b-c)^2+4bc`

`color(white)((a^2-4)^2) = (-9/a)^2+144`

That is:

`a^4-8a^2+16 = 81/a^2+144`

Multiplying both sides by `a^2` and rearranging slightly, this becomes:

`(a^2)^3-8(a^2)^2-128(a^2)-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(a^2)^3-216(a^2)^2-3456(a^2)-2187`

`color(white)(0) = (3a^2-8)^3-1344(3a-8)-12427`

`color(white)(0) = t^3-1344t-12427`

where `t=3a^2-8`

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

Let:

`t = 16sqrt(7) cos theta`

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

Then:

`0 = t^3-1344t-12427`

`color(white)(0) = 7168 sqrt(7) (4 cos^3 theta-3 cos theta) - 12427`

`color(white)(0) = 7168 sqrt(7) cos 3 theta - 12427`

So:

`cos 3 theta = 12427/(7168 sqrt(7)) = 12427/50176 sqrt(7)`

Hence:

`3 theta = +-cos^(-1)(12427/50176 sqrt(7))+2npi`

So:

`theta = +-1/3cos^(-1)(12427/50176 sqrt(7))+(2npi)/3`

This gives distinct solutions:

`t_n = 16sqrt(7)cos(1/3cos^(-1)(12427/50176 sqrt(7))+(2npi)/3)`

for `n = 0, 1, 2`

The one we are interested in is `t_0 ~~ 40.6194` since it is the only positive one.

Then we can choose:

`a = sqrt(1/3(8+16sqrt(7)cos(1/3cos^(-1)(12427/50176 sqrt(7)))))`

Then:

`b = 1/2(a^2-4-9/a)`

`c = 1/2(a^2-4+9/a)`

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

`f^2-af+b`

`f^2+af+c`

The details get very messy.