2007 AMC 12A Problems/Problem 18

Revision as of 18:43, 27 October 2013 by Armalite46 (talk | contribs) (Solution)

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

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 12 Problems and Solutions

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