Question

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

Polynomial factorization
ANSWERED
asked 2021-05-12
Why isn't factoring \(81x^{4}-16\) as \((9x^{2}+4)(9x^{2}-4)\) a complete factorization?

Answers (1)

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:
\(81x^{4}-16=(9x^{2}+4)(9x^{2}-4)\)
\(=(9x^{2}+4)(3x-2)(3x+2)\)
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