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 $-39$.
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 $-39$.
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.