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!
