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

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Answer 1 · 2016-03-25T00:06:55

$4q^2+27r^4=(2q-3sqrt(3)ir^2)(2q+3sqrt(3)ir^2)$

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

We can do something with Complex coefficients.

I will use the difference of squares identity:

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

with $a = 2q$ and $b=3sqrt(3)ir^2$

$4q^2+27r^4$

$=(2q)^2+(sqrt(27)r^2)^2$

$=(2q)^2+(3sqrt(3)r^2)^2$

$=(2q)^2-(3sqrt(3)ir^2)^2$

$=(2q-3sqrt(3)ir^2)(2q+3sqrt(3)ir^2)$

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

How do you factor the expression 4q^2 + 27r^44q^2 + 27r^4?