Answer:
The correct answer is B,
Step-by-step explanation:
Since the higher value of z's power, is 4, an even number, and the second number is negative, we can think that this binomial is made of a difference of squares, so that is what we are going to factorize.
First, extract the common number (if any), the number 2, we have now>
[tex]2*(z^{4}-1)[/tex]
This is convenient since "1" is a wonderful number that has this feature>[tex]1^{n}= 1[/tex] No matters what n's value is,
so the first equation [tex]2*(z^{4}-1)[/tex] can be written as [tex](2*(z^{4}-1)=2*((z^{2})^{2}-1^{2})=2*(z^{2}-1)*(z^{2}+1)[/tex]
The later termn, can also be factorized, using the same as befre.
[tex](z^{2}-1)=(z-1)*(z+1)[/tex]
Remember that [tex]z^{4}=(z^{2})^{2}[/tex]
