How do you simplify #(2z^3+128)/(16+8z+z^2)#?

Question

Eli Assouline · Accepted Answer

The expression can be simplified to #(2(z^2 - 4z + 16))/(z + 4)# with a restriction of #z !=-4#.

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#. Thus, our original statement 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#. I hope this is useful!

Isaiah Anthony · Answer

undefined To simplify the expression (2z^3+128)/(16+8z+z^2), we can factor the numerator and denominator and then cancel out any common factors.

The numerator can be factored as 2(z^3+64), and the denominator can be factored as (z+8)^2.

Now, we can cancel out the common factor of 2 and simplify further.

The simplified expression is (z^3+64)/(8+z).