How do you factor #f^4-4f^2-9f+36#?
2 Answers
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)#
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^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)^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
#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
This cubic in
Let:
#t = 16sqrt(7) cos theta#
(The multiplier
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
The one we are interested in is
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.