AMC12 2017 A

AMC12 2017 A · Q23

AMC12 2017 A · Q23. It mainly tests Systems of equations, Polynomials.

For certain real numbers $a$, $b$, and $c$, the polynomial $g(x) = x^3 + a x^2 + x + 10$ has three distinct roots, and each root of $g(x)$ is also a root of the polynomial $f(x) = x^4 + x^3 + b x^2 + 100 x + c$. What is $f(1)$?

对于某些实数 $a$、$b$ 和 $c$，多项式 $g(x) = x^3 + a x^2 + x + 10$ 有三个不同根，且 $g(x)$ 的每个根也是多项式 $f(x) = x^4 + x^3 + b x^2 + 100 x + c$ 的根。求 $f(1)$？

(A) -9009 -9009

(B) -8008 -8008

(D) -6006 -6006

(E) -5005 -5005

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $q$ be the additional root of $f(x)$. Then $$ f(x)=(x-q)(x^3+ax^2+x+10) $$ $$ =x^4+(a-q)x^3+(1-qa)x^2+(10-q)x-10q. $$ Thus $100=10-q$, so $q=-90$ and $c=-10q=900$. Also $1=a-q=a+90$, so $a=-89$. It follows, using the factored form of $f$ shown above, that $$ f(1)=(1-(-90))(1-89+1+10)=91\cdot(-77)=-7007. $$

答案（C）：设 $q$ 为 $f(x)$ 的另一个根，则 $$ f(x)=(x-q)(x^3+ax^2+x+10) $$ $$ =x^4+(a-q)x^3+(1-qa)x^2+(10-q)x-10q. $$ 因此 $100=10-q$，所以 $q=-90$ 且 $c=-10q=900$。又有 $1=a-q=a+90$，所以 $a=-89$。由上面的 $f$ 的因式分解形式可得 $$ f(1)=(1-(-90))(1-89+1+10)=91\cdot(-77)=-7007. $$

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