How do you factor $36 a^{4} + 90 a^{2} b^{2} + 99 b^{4}$ $36a^4 + 90a^2b^2 + 99b^4$ ?

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How do you factor $36 a^{4} + 90 a^{2} b^{2} + 99 b^{4}$ $36a^4 + 90a^2b^2 + 99b^4$ ?

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Answer 1 · 2016-05-12T21:38:59

$36 a^{4} + 90 a^{2} b^{2} + 99 b^{4}$ $36a^4+90a^2b^2+99b^4$

$= (6 a^{2} - 3 (\sqrt{4 \sqrt{11} - 10}) a b + 3 \sqrt{11} b^{2}) (6 a^{2} + 3 (\sqrt{4 \sqrt{11} - 10}) a b + 3 \sqrt{11} b^{2})$ $=(6a^2-3(sqrt(4sqrt(11)-10))ab+3sqrt(11)b^2)(6a^2+3(sqrt(4sqrt(11)-10))ab+3sqrt(11)b^2)$

Explanation:

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

$(6 a^{2} - k a b + 3 \sqrt{11} b^{2}) (6 a^{2} + k a b + 3 \sqrt{11} b^{2})$ $(6a^2-kab+3sqrt(11)b^2)(6a^2+kab+3sqrt(11)b^2)$

$= 36 a^{4} + (36 \sqrt{11} - k^{2}) a^{2} b^{2} + 99 b^{4}$ $=36a^4+(36sqrt(11)-k^2)a^2b^2+99b^4$

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

$90 = 36 \sqrt{11} - k^{2}$ $90 = 36sqrt(11)-k^2$

Add $k^{2} - 90$ $k^2-90$ to both sides to get:

$k^{2} = 36 \sqrt{11} - 90$ $k^2 = 36sqrt(11)-90$

So:

$k = \pm \sqrt{36 \sqrt{11} - 90} = \pm 3 \sqrt{4 \sqrt{11} - 10}$ $k = +-sqrt(36sqrt(11)-90) = +-3sqrt(4sqrt(11)-10)$

So:

$36 a^{4} + 90 a^{2} b^{2} + 99 b^{4}$ $36a^4+90a^2b^2+99b^4$

$= (6 a^{2} - 3 (\sqrt{4 \sqrt{11} - 10}) a b + 3 \sqrt{11} b^{2}) (6 a^{2} + 3 (\sqrt{4 \sqrt{11} - 10}) a b + 3 \sqrt{11} b^{2})$ $=(6a^2-3(sqrt(4sqrt(11)-10))ab+3sqrt(11)b^2)(6a^2+3(sqrt(4sqrt(11)-10))ab+3sqrt(11)b^2)$

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How do you factor $36 a^{4} + 90 a^{2} b^{2} + 99 b^{4}$ $36a^4 + 90a^2b^2 + 99b^4$ ?

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