How do you simplify 2z3+12816+8z+z2?

1 Answer
Noah G
Oct 2, 2016

The expression can be simplified to 2(z2−4z+16)z+4 with a restriction of z≠−4.

Explanation:

Factor

=2(z3+64)(z+4)(z+4)

Use synthetic division to factor the expression z3+64. We know that z+4 is a factor, because by the remainder theorem f(−4)=(−4)3+64=0, if f(x)=z3+64.

−4_|1 0 0 64
−4 16 −64
--------------------------------------------------
1 −4 16 0

Hence, when z3+64 is divided by z+4, the quotient is z2−4z+16 with a remainder of 0. The expression z2−4z+16 is not factorable, however, because no two numbers multiply to +16 and add to −4.

So, our initial expression becomes:

=2(z+4)(z2−4z+16)(z+4)(z+4)

Now, eliminate using the property aa=1,a≠0

=2(z2−4z+16)z+4

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

z2+8x+16=0

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

z=−4

Hence, z≠−4.

Hopefully this helps!

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