Question

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

Answer 1

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

Explanation:

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!