How do you solve $2x^3 + 4x^2 + 4x + 8$ when x = 15?

Question

1868 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-26T06:29:25

The answer is $7718$ .

Substitute $15$ for $x$ .

$2x^3+4x^2+4x+8$

$2(15^3)+4(15^2)+4(15)+8=$

$6750+900+60+8=7718$

How do you solve 2x^3 + 4x^2 + 4x + 82x^3 + 4x^2 + 4x + 8 when x = 15?