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Suppose $$f(x) = ax^2 + bx + c$$. $$f(x)$$ leaves a remainder 18 when divided by $$x + 2$$ and a remainder 13 when divided by $$x - 3$$. Also, $$f(0) = 4$$. Then $$2a + b + c$$ is equal to
By the remainder theorem $$f(-2) = 4a - 2b + c = 18$$, $$f(3) = 9a + 3b + c = 13$$ and $$c = f(0) = 4$$. So $$2a - b = 7$$ and $$3a + b = 3$$, giving $$a = 2$$ and $$b = -3$$. Hence $$2a + b + c = 4 - 3 + 4 = 5$$.
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