Factor

#=(2(z^3 + 64))/((z + 4)(z + 4))#

Use synthetic division to factor the expression #z^3 + 64#. We know that #z + 4# is a factor, because by the remainder theorem #f(-4) = (-4)^3 + 64 = 0#, if #f(x) = z^3 + 64#.

#-4"_|"1" "0" "0" "64"#

#" " -4" "16" "-64"#

#"--------------------------------------------------"#

#" "1" " -4" "16" "0#

Hence, when #z^3 + 64# is divided by #z + 4#, the quotient is #z^2 - 4z + 16# with a remainder of #0#. The expression #z^2 - 4z + 16# is not factorable, however, because no two numbers multiply to #+16# and add to #-4#.

So, our initial expression becomes:

#=(2(z + 4)(z^2 - 4z + 16))/((z + 4)(z + 4))#

Now, eliminate using the property #a/a = 1, a != 0#

#=(2(z^2 - 4z + 16))/(z + 4)#

Finally, state your restrictions on the variable. This can be done by setting the **original** expression to #0# and solving.

#z^2 + 8x+ 16 = 0#

#(z + 4)(z + 4) = 0#

#z = -4#

Hence, #z!=-4#.

Hopefully this helps!