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2007 AMC 12A Problem 18

Problem 18 of 25IntermediateAlgebra

The polynomial f(x)=x4+ax3+bx2+cx+df(x)=x^4+ax^3+bx^2+cx+d has real coefficients, and f(2i)=f(2+i)=0.f(2i)=f(2+i)=0. What is a+b+c+d?a+b+c+d?

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Solution

Since ff has real coefficients, the conjugates −2i-2i and 2−i2-i are also roots. Thus f(x)=(x2+4)(x2−4x+5)=x4−4x3+9x2−16x+20. \begin{gathered} f(x) = (x^2+4)(x^2-4x+5) \\ = x^4-4x^3+9x^2 \\ {}-16x+20. \end{gathered} Then a+b+c+d=−4+9a+b+c+d=-4+9 −16+20-16+20 =9.=9. Equivalently, a+b+c+d=f(1)−1a+b+c+d=f(1)-1 =(1+4)(1+1)−1=(1+4)(1+1)-1 =9.=9. Thus, the correct answer is D.

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Concepts: complex number · polynomial

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