How do you factor the trinomial #9a^5b + 33ab^5#?

1 Answer
George C.
Nov 18, 2015

#9a^5b+33ab^5= 3ab(3a^4+11b^4)#

#= 3ab(sqrt(3)a^2+root(4)(132)ab+sqrt(11)b^2)(sqrt(3)a^2-root(4)(132)ab+sqrt(11)b^2)#

Explanation:

This is actually a binomial (having two terms), not a trinomial (three terms).

That aside, first we identify that both of the terms are divisible by #3ab#, so separate that factor out first:

#9a^5b+33ab^5= 3ab(3a^4+11b^4)#

Since the coefficients of the quartic are both positive, this has no linear factors with Real coefficients, but it does have quadratic factors with Real coefficients:

Consider:

#(c^2+2cd+2d^2)(c^2-2cd+2d^2)=c^4+4d^4#

If we let #c = root(4)(3)a# and #d = root(4)(11/4)b# then:

#3a^4+11b^4#

#= c^4+4d^4#

#=(c^2+2cd+2d^2)(c^2-2cd+2d^2)#

#=(sqrt(3)a^2+2root(4)(33/4)ab+2sqrt(11/4)b^2)(sqrt(3)a^2-2root(4)(33/4)ab+2sqrt(11/4)b^2)#

#=(sqrt(3)a^2+root(4)(132)ab+sqrt(11)b^2)(sqrt(3)a^2-root(4)(132)ab+sqrt(11)b^2)#

Putting this all together:

#9a^5b+33ab^5#

#= 3ab(sqrt(3)a^2+root(4)(132)ab+sqrt(11)b^2)(sqrt(3)a^2-root(4)(132)ab+sqrt(11)b^2)#

Footnote

The identity:

#(c^2+2cd+2d^2)(c^2-2cd+2d^2)=c^4+4d^4#

came from thinking about #4#th roots of #-1#

#-1# has square roots #+-i#, which in turn have square roots:

#1/sqrt(2) (+-1+-i)#

Rather than mess about with a #1/sqrt(2)# factor, we can look at #root(4)(-4)# instead.

#-4# has square roots #+-2i#, which in turn have square roots:

#+-1+-i#

So:

#x^4+4 = (x+1-i)(x+1+i)(x-1-i)(x-1+i)#

#=((x+1)^2+1)((x-1)^2+1)#

#=(x^2+2x+2)(x^2-2x+2)#

Hence:

#c^4+4d^4 = (c^2+2cd+2d^2)(c^2-2cd+2d^2)#

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