How do you factor #36a^4 + 90a^2b^2 + 99b^4#?

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Answer 1 · 2016-05-12T21:38:59

#36a^4+90a^2b^2+99b^4#

#=(6a^2-3(sqrt(4sqrt(11)-10))ab+3sqrt(11)b^2)(6a^2+3(sqrt(4sqrt(11)-10))ab+3sqrt(11)b^2)#

Taking square roots of the first and last term, let us try a factorisation of the form:

#(6a^2-kab+3sqrt(11)b^2)(6a^2+kab+3sqrt(11)b^2)#

#=36a^4+(36sqrt(11)-k^2)a^2b^2+99b^4#

So all we need to do is choose #k# such that:

#90 = 36sqrt(11)-k^2#

Add #k^2-90# to both sides to get:

#k^2 = 36sqrt(11)-90#

So:

#k = +-sqrt(36sqrt(11)-90) = +-3sqrt(4sqrt(11)-10)#

So:

#36a^4+90a^2b^2+99b^4#

#=(6a^2-3(sqrt(4sqrt(11)-10))ab+3sqrt(11)b^2)(6a^2+3(sqrt(4sqrt(11)-10))ab+3sqrt(11)b^2)#