How do you factor f^4-4f^2-9f+36f4−4f2−9f+36?
2 Answers
Explanation:
Assuming a typo in the question, suppose we want to factor:
f^color(red)(3)-4f^2-9f+36f3−4f2−9f+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)f3−4f2−9f+36=(f3−4f2)−(9f−36)
color(white)(f^3-4f^2-9f+36) = f^2(f-4)-9(f-4)f3−4f2−9f+36=f2(f−4)−9(f−4)
color(white)(f^3-4f^2-9f+36) = (f^2-9)(f-4)f3−4f2−9f+36=(f2−9)(f−4)
color(white)(f^3-4f^2-9f+36) = (f^2-3^2)(f-4)f3−4f2−9f+36=(f2−32)(f−4)
color(white)(f^3-4f^2-9f+36) = (f-3)(f+3)(f-4)f3−4f2−9f+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)f4−4f2−9f+36=(f2−af+b)(f2+af+c)
color(white)(f^4-4f^2-9f+36) = f^4+(b+c-a^2)f^2+a(b-c)f+bcf4−4f2−9f+36=f4+(b+c−a2)f2+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.