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:

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

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

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:

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

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