Question

Factor completely 21b^4 +5b^2m^2- 4m^2

Answer 1

Factor completely

$21b^4+5b^2m^2-4m^4$

We need to split the expression into convenient groups

$21b^4+5b^2m^2-4m^4=21b^4+12b^2m^2-4m^4-7b^2m^2$

Factor 3b^2 from the first group and -m^2 from the second group:

$21b^4+5b^2m^2-4m^4=3b^2(7b^2+4m^2)-m^2(4m^2+7b^2)$

Factor out the common term 7b^2+4m^2:

$21b^4+5b^2m^2-4m^4=(3b^2-m^2)(4m^2+7b^2)$

This is the final factorization