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Algebra Difficulty 4.9 AIME Find the answer

Find the sum of the coefficients of the polynomial P(x)=x429x3+ax2+bx+cP(x)=x^{4}-29 x^{3}+a x^{2}+b x+c, given that P(5)=11,P(11)=17P(5)=11, P(11)=17, and P(17)=23P(17)=23.

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Solution

Define Q(x)=P(x)x6=x429x3+ax2+(b1)x+(c6)Q(x)=P(x)-x-6=x^{4}-29 x^{3}+a x^{2}+(b-1)x+(c-6) and notice that Q(5)=Q(11)=Q(17)=0Q(5)=Q(11)=Q(17)=0. Q(x)Q(x) has degree 4 and by Vieta's Formulas the sum of its roots is 29, so its last root is 2917115=429-17-11-5=-4, giving us Q(x)=(x5)(x11)(x17)(x+4)Q(x)=(x-5)(x-11)(x-17)(x+4). This means that P(1)=Q(1)+7=(4)(10)(16)(5)+7=3200+7=3193P(1)=Q(1)+7=(-4)(-10)(-16)(5)+7=-3200+7=-3193.

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