Question

# Why isn't factoring 81x^{4}-16 as (9x^{2}+4)(9x^{2}-4) a complete factorization?

Polynomial factorization
Why isn't factoring $$81x^{4}-16$$ as $$(9x^{2}+4)(9x^{2}-4)$$ a complete factorization?

2021-05-13
Step 1
Given, why isn't factoring
$$81x^{4}-16\ as\ (9x^{2}+4)(9x^{2}-4)$$
a complete factorization?
Step 2
Consider,
$$(9x^{2}+4)(9x^{2}-4)$$ this term can be again factored as
$$(9x^{2}+4)(9x^{2}-4) = (9x^{2}+4)(3x-2)(3x+2)$$
So, the complete factorization of $$81x^{4}-16$$ is as follows:
$$81x^{4}-16=(9x^{2}+4)(9x^{2}-4)$$
$$=(9x^{2}+4)(3x-2)(3x+2)$$