How do you factor #y^4+8y^3+12#?

1 Answer
Feb 20, 2017

Answer:

If the question is correct as stands then the answer is complicated, but here's a sketch of how you can do it...

Explanation:

#y^4+8y^3+12#

Let #t = y+2#

Then:

#y^4+8y^3+12 = (y+2)^4-24(y+2)^2+64(y+2)-36#

#color(white)(y^4+8y^3+12) = t^4-24t^2+64t-36#

#color(white)(y^4+8y^3+12) = (t^2-at+b)(t^2+at+c)#

#color(white)(y^4+8y^3+12) = t^4+(b+c-a^2)t^2+a(b-c)t+bc#

Equating coefficients:

#{ (b+c = a^2-24), (b-c = 64/a), (bc=-36) :}#

Hence:

#a^4-48a^2+576 = (a^2-24)^2 = (b+c)^2 = (b-c)^2+4bc = 4096/a^2-144#

Hence:

#0 = (a^2)^3-48(a^2)^2+720(a^2)-4096#

#color(white)(0) = (a^2-16)^3-48(a^2-16)-768#

Using Cardano's method, let #a^2-16 = u+v#

Then:

#u^3+v^3+3(uv-16)(u+v)-768 = 0#

Add the constraint #v = 16/u# to eliminate the term in #(u+v)# and get:

#(u^3)^2-768(u^3)+4096 = 0#

Using the quadratic formula:

#u^3 = (768+-sqrt(768^2-4(1)(4096)))/2#

#color(white)(u^3) = 384+-64sqrt(35)#

#color(white)(u^3) = 4^3(6+-sqrt(35))#

Hence:

#a^2 = 16+4root(3)(6+sqrt(35))+4root(3)(6-sqrt(35))#

We can use the positive square root:

#a = 2sqrt(4+root(3)(6+sqrt(35))+root(3)(6-sqrt(35)))#

Then:

#b = 1/2(a^2-24+64/a)#

#c = 1/2(a^2-24-64/a)#

Leaving us with two quadratics to solve:

#t^2-at+b = 0#

#t^2+at+c = 0#

yielding #4# roots: #t_1, t_2, t_3, t_4#.

Then the zeros of our original quartic are given by:

#y_n = t_n-2#

Hence we find a factorisation:

#y^4+8y^3+12 = (y-y_1)(y-y_2)(y-y_3)(y-y_4)#

where #y_1, y_2, y_3, y_4# are the four zeros.