18) Let $a, b, c$ be the solutions of the equation $x^{3}-3 x^{2}-18 x+40=0$. Knowing that $a b=10$, calculate $c(a+b)$.
(C) 21
(D) 22
(E) cannot be determined.
18) The answer is (A)

Let $f(x)=x^{3}-3 x^{2}-18 x+40$. Since $x^{3}-3 x^{2}-18 x+40=(x-a)(x-b)(x-c)=$ $x^{3}-(a+b+c) x^{2}+(a b+a c+b c)_{x}-a b c$, we have $a b c=-40$. Since $a b=10$, it follows that $c=-4$. Furthermore, $a+b+c=3$, from which $a+b=7$ and $c(a+b)=(-4) \times 7=-28$.

## SECOND SOLUTION.

Using the previous factorization, we obtain $a b+a c+b c=10+c(a+b)=-18$, from which $c(a+b)=-28$.