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

2 Answers
Jan 14, 2018

Answer:

#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)#

Jan 14, 2018

Answer:

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.