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

Question

Hazel Anglin · Accepted Answer

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

#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.

Eva Adamson · Answer

undefined To factor $y^4 + 8y^3 + 12$, you can rewrite it as $(y^4 + 8y^3) + 12$. Then, factor out the greatest common factor from the first two terms, which is $y^3$, leaving you with $y^3(y + 8) + 12$.