2007 AMC 12A Problems/Problem 18

Problem

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

$\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 4 \qquad \mathrm{(D)}\ 9 \qquad \mathrm{(E)}\ 16$

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, namely $2-i,-2i$ . Now we work backwards for the polynomial:

$(x-(2+i))(x-(2-i))(x-2i)(x+2i) = 0$

$(x^2 - 4x + 5)(x^2 + 4) = 0$

$x^4 - 4x^3 + 9x^2 - 16x + 20 = 0$

Thus our answer is $- 4 + 9 - 16 + 20 = 9\ \mathrm{(D)}$ .

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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2007 AMC 12A Problems/Problem 18

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