How do you factor f^4-4f^2-9f+36f44f29f+36?

2 Answers
Jan 14, 2018

f^color(red)(3)-4f^2-9f+36 = (f-3)(f+3)(f-4)f34f29f+36=(f3)(f+3)(f4)

Explanation:

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

f^color(red)(3)-4f^2-9f+36f34f29f+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)f34f29f+36=(f34f2)(9f36)

color(white)(f^3-4f^2-9f+36) = f^2(f-4)-9(f-4)f34f29f+36=f2(f4)9(f4)

color(white)(f^3-4f^2-9f+36) = (f^2-9)(f-4)f34f29f+36=(f29)(f4)

color(white)(f^3-4f^2-9f+36) = (f^2-3^2)(f-4)f34f29f+36=(f232)(f4)

color(white)(f^3-4f^2-9f+36) = (f-3)(f+3)(f-4)f34f29f+36=(f3)(f+3)(f4)

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^3f3 and hence there is a factorisation of the form:

f^4-4f^2-9f+36 = (f^2-af+b)(f^2+af+c)f44f29f+36=(f2af+b)(f2+af+c)

color(white)(f^4-4f^2-9f+36) = f^4+(b+c-a^2)f^2+a(b-c)f+bcf44f29f+36=f4+(b+ca2)f2+a(bc)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.