How do you factor the expression $4 q^{2} + 27 r^{4}$ $4q^2 + 27r^4$ ?

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How do you factor the expression $4 q^{2} + 27 r^{4}$ $4q^2 + 27r^4$ ?

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Answer 1 · 2016-03-25T00:06:55

$4 q^{2} + 27 r^{4} = (2 q - 3 \sqrt{3} i r^{2}) (2 q + 3 \sqrt{3} i r^{2})$ $4q^2+27r^4=(2q-3sqrt(3)ir^2)(2q+3sqrt(3)ir^2)$

Explanation:

Since both cofficients are positive and the one in $q$ $q$ is of degree $2$ $2$ , this has no simpler polynomial factors with Real coefficients.

We can do something with Complex coefficients.

I will use the difference of squares identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

with $a = 2 q$ $a = 2q$ and $b = 3 \sqrt{3} i r^{2}$ $b=3sqrt(3)ir^2$

$4 q^{2} + 27 r^{4}$ $4q^2+27r^4$

$= {(2 q)}^{2} + {(\sqrt{27} r^{2})}^{2}$ $=(2q)^2+(sqrt(27)r^2)^2$

$= {(2 q)}^{2} + {(3 \sqrt{3} r^{2})}^{2}$ $=(2q)^2+(3sqrt(3)r^2)^2$

$= {(2 q)}^{2} - {(3 \sqrt{3} i r^{2})}^{2}$ $=(2q)^2-(3sqrt(3)ir^2)^2$

$= (2 q - 3 \sqrt{3} i r^{2}) (2 q + 3 \sqrt{3} i r^{2})$ $=(2q-3sqrt(3)ir^2)(2q+3sqrt(3)ir^2)$

This is as far as we can go, even with Complex coefficients, since the degree of the $2 q$ $2q$ terms is $1$ $1$ .

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