AMC10 2017 A

AMC10 2017 A · Q24

AMC10 2017 A · Q24. It mainly tests Polynomials, Algebra misc.

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 $g(x) = x^3 + a x^2 + x + 10$ 有三个不同的根，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 $f(1)=(1-(-90))\cdot(1-89+1+10)=91\cdot(-77)=-7007.$

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 $a=-89$。由上面给出的 $f$ 的因式分解形式可得 $f(1)=(1-(-90))\cdot(1-89+1+10)=91\cdot(-77)=-7007。$

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