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What is shortcut to this contest algebra problem about polynomial?

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 $c$.

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marked as duplicate by Srivatsan, Dylan Moreland, t.b., Jonas Teuwen, Asaf Karagila Dec 20 '11 at 20:36

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Hint: What can you say about root of the polynomial $P(x)-11x$? How are the roots related to coefficients? –  Martin Sleziak Dec 20 '11 at 19:45

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$P(x)-11x=(x-1)(x-2)(x-3)(x-4)$ since it is monic, has degree four, and is zero at $1,2,3,4$. the coefficient of $x$ is $-(1\cdot2\cdot3+1\cdot2\cdot4+1\cdot3\cdot4+2\cdot3\cdot4)=-(6+8+12+24)=-50$. hence the coefficient of $x$ in $P(x)$ is $c=-39$. likewise, $a=-(4+3+2+1)=-10$, $b=1\cdot2+1\cdot3+1\cdot4+2\cdot3+2\cdot4+3\cdot4=2+3+4+6+8+12=35$ and $d=1\cdot2\cdot3\cdot4=24$ so that $P(x)=x^4-10x^3+35x^2-39x+24$

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"since it has degree four and is zero at 1,2,3,4" and is monic. –  Jason DeVito Dec 20 '11 at 20:28

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