AMC12 2007 A

AMC12 2007 A · Q18

AMC12 2007 A · Q18. It mainly tests Polynomials, Complex numbers (rare).

The polynomial $f(x) = x^{4} + ax^{3} + bx^{2} + cx + d$ has real coefficients, and $f(2i) = f(2 + i) = 0.$ What is $a + b + c + d?$

多项式 $f(x) = x^{4} + ax^{3} + bx^{2} + cx + d$ 有实系数，且 $f(2i) = f(2 + i) = 0.$ 求 $a + b + c + d$？

(A) 0 0

(B) 1 1

(D) 9 9

(E) 16 16

Answer

Correct choice: (D)

正确答案：（D）

Solution

A fourth degree polynomial has four roots. Since the coefficients are real(meaning that complex roots come in conjugate pairs), the remaining two roots must be the complex conjugates of the two given roots. By the factor theorem, our roots are $2-i,-2i$. Now we work backwards for the polynomial: Thus our answer is $- 4 + 9 - 16 + 20 = 9\ \mathrm{(D)}$.

因为 $f(x)$ 有实系数，且 $2i$ 和 $2 + i$ 是根，所以它们的共轭 $-2i$ 和 $2 - i$ 也是根。因此 $f(x) = (x + 2i)(x - 2i)(x - (2 + i))(x - (2 - i)) = (x^2 + 4)(x^2 - 4x + 5)$ $= x^4 - 4x^3 + 9x^2 - 16x + 20$。因此 $a + b + c + d = -4 + 9 - 16 + 20 = 9$。

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