20) The polynomial $x^{4}-2 x^{3}-7 x^{2}-2 x+1$ has four real roots $a, b, c, d$. What is the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$?
(A) -2 (B) 0
(C) -1
(D) 2
(E) none of the above values.
20) The answer is (D). First, observe that if $\alpha$ is a root of the given polynomial, then $\frac{1}{\alpha}$ is also a root. Indeed, $\alpha$ is different from 0, and dividing by $\alpha^{4}$ yields precisely $1-\frac{2}{\alpha}-\frac{7}{\alpha^{2}}-\frac{2}{\alpha^{3}}+\frac{1}{\alpha^{4}}=0$. (This calculation could have been avoided by noting that the polynomial is palindromic). Therefore, the set of reciprocals of the roots coincides with the set of the roots, and thus the sum of the roots is equal to the sum of their reciprocals. But the sum of the roots of a monic polynomial is the opposite of the second coefficient.

SECOND SOLUTION.

If $a, b, c, d$ are the roots of the given polynomial, then

$$
(x-a)(x-b)(x-c)(x-d)=x^{4}-2 x^{3}-7 x^{2}-2 x+1,
$$

so $1=a b c d$ and $2=a b c+a b d+a c d+b c d$.

On the other hand,

$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{b c d+a c d+a b d+a b c}{a b c d} .
$$

Therefore, the sum of the reciprocals of the roots is 2.