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The polynomial $P(x)=x^4 + ax^3 + bx^2 +cx + d$ has the property that $p(k)=11k$ for $k=1,2,3,4$. Compute $c$.

The answer is $-39$.

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A little rude to be dictating how people should help you with their time by giving you answers to your questions. – Arturo Magidin Dec 8 '11 at 2:41
@ArturoMagidin - i think a short solution would help any math-blind person like me. – Victor Dec 8 '11 at 2:44
I'm sure you do. But it's still rude to try to dictate to people when you are asking for their help. Limosneros con garrote often find themselves with no help. – Arturo Magidin Dec 8 '11 at 2:46
@Arturo: There is apparently a movie with the same title! – Quixotic Dec 8 '11 at 2:50
@MaX: It's a common phrase, at least in Mexico. – Arturo Magidin Dec 8 '11 at 3:10
up vote 14 down vote accepted

Let $Q(x)=P(x)-11x$. Then the roots of $Q(x)$ are $1$, $2$, $3$, and $4$.

But if the roots of a monic quartic are $r_1, r_2, r_3, r_4$, then the sum of the products of the $r_i$, taken $3$ at a time, is the negative of the coefficient of $x$. For $1,2,3,4$ this sum of products is $50$. A simple way to calculate is to divide $24$ by $1$, $2$, $3$, and $4$ and add up the results.

We conclude that the coefficient of $x$ in $Q(x)$ is $-50$. This is $c-11$, so $c=-39$.

There may be a simpler approach, but probably this essentially automatic use of one of the Viète relations is the intended argument.

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