AMC12 2021 A

AMC12 2021 A · Q12

AMC12 2021 A · Q12. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$?

多项式$z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$的所有根都是正整数，可能有重根。$B$的值是多少？

(A) {-}88 {-}88

(B) {-}80 {-}80

(D) {-}41 {-}41

(E) {-}40 {-}40

Answer

Correct choice: (A)

正确答案：（A）

Solution

By Vieta's formulas, the sum of the six roots is $10$ and the product of the six roots is $16$. By inspection, we see the roots are $1, 1, 2, 2, 2,$ and $2$, so the function is $(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$. Therefore, calculating just the $z^3$ terms, we get $B = -32 - 48 - 8 = \boxed{\textbf{(A) }{-}88}$.

由Vieta公式，六个根的和为$10$，六个根的积为$16$。通过检查，我们看到根为$1, 1, 2, 2, 2,$ 和 $2$，因此函数为$(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$。因此，计算仅$z^3$项，我们得到$B = -32 - 48 - 8 = \boxed{\textbf{(A) }{-}88}$。

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