Question

What is the completely factored form of t4 − 16?

Answer 1

we know that

A difference of squares problem can be factored as follows:

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

in this problem we have

$t^{4}-16$

Applying difference of squares

$t^{4}-16=(t^{2}-4)( t^{2}+4)$

the term

$(t^{2}-4)=(t+2)(t-2)$

substitute

$(t^{2}-4)( t^{2}+4)=(t+2)(t-2)( t^{2}+4)$

therefore

the answer is

$(t+2)(t-2)( t^{2}+4)$