well, the remainder theorem states that for an expression of x + a dividing a polynomial say f(x), will gives some remainder, well that remainder is exactly the same value as if we simply just take f(-a).
so, let's take a peek, we're given f(-3), that means that x = -3, which we can write as x + 3 = 0.
that means that if we divide f(x) by x + 3, we'll get the same value as if we simply solve for f(-3), let's check, let's divide 2x² - 5x - 8 by x + 3, using synthetic division.
![\bf x + 3 = 0\implies x = \boxed{-3}~~\leftarrow \textit{our divisor for the synthetic division} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{r|rrrr} -3&2&-5&-8\\ &&-6&33&\\\cline{1-5} &2&-11&\underline{25}&\leftarrow remainder \end{array} \\\\[-0.35em] ~\dotfill\\\\ f(x)=2x^2-5x-8\implies f(-3)=2(-3)^2-5(-3)-8\implies f(-3)=\underline{25}](https://tex.z-dn.net/?f=%5Cbf%20x%20%2B%203%20%3D%200%5Cimplies%20x%20%3D%20%5Cboxed%7B-3%7D~~%5Cleftarrow%20%5Ctextit%7Bour%20divisor%20for%20the%20synthetic%20division%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Br%7Crrrr%7D%20-3%262%26-5%26-8%5C%5C%20%26%26-6%2633%26%5C%5C%5Ccline%7B1-5%7D%20%262%26-11%26%5Cunderline%7B25%7D%26%5Cleftarrow%20remainder%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20f%28x%29%3D2x%5E2-5x-8%5Cimplies%20f%28-3%29%3D2%28-3%29%5E2-5%28-3%29-8%5Cimplies%20f%28-3%29%3D%5Cunderline%7B25%7D)