Using the remainder theorem to solve this equation?

P(x) = 2x^4 - 5x^3 - 5x^2 + 8
P(3) = 2(3)^4 - 5(3)^3 - 5(3)^2 + 8
P(3) = - 10 answer//


3|....2....- 5...- 5....0....8
..............6.....3..-6...- 18
.-------------------------------------
.....2........1...-2..- 6...- 10 < remainder//

By long division, or whatever...
2 1 -2 -6
x-3 / 2 -5 -5 0 8
2 -6
-------------
1 -5
1 -3
-------------
-2 0 8
-2 6
--------
-6 8
-6 18
-----
-10
You will find that
2x^4 - 5x^3 - 5x^2 + 8 = (x-3)(2x^3+x^2-2x-6) - 10
[Quotient = 2x^3+x^2-2x-6, reminder = -10]
Therefore, by reminder theorem, P(3) = -10.