Difference between revisions of "2007 AMC 12A Problems/Problem 18"

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The [[polynomial]] <math>f(x) = x^{4} + ax^{3} + bx^{2} + cx + d</math> has real [[coefficient]]s, and <math>f(2i) = f(2 + i) = 0.</math> What is <math>a + b + c + d?</math>
 
The [[polynomial]] <math>f(x) = x^{4} + ax^{3} + bx^{2} + cx + d</math> has real [[coefficient]]s, and <math>f(2i) = f(2 + i) = 0.</math> What is <math>a + b + c + d?</math>
 
 
<math>\displaystyle \mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 4 \qquad \mathrm{(D)}\ 9 \qquad \mathrm{(E)}\ 16</math>
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<math>\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 4 \qquad \mathrm{(D)}\ 9 \qquad \mathrm{(E)}\ 16</math>
  
 
==Solution==
 
==Solution==
A fourth degree polynomial has four [[root]]s. Since the coefficients are real, the remaining two roots must be the [[complex conjugate]]s of the two given roots, namely <math>2+i, \displaystyle -2i</math>. Now we work backwards for the polynomial:
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A fourth degree polynomial has four [[root]]s. Since the coefficients are real, the remaining two roots must be the [[complex conjugate]]s of the two given roots, namely <math>2+i,-2i</math>. Now we work backwards for the polynomial:
  
 
<div style="text-align:center;"><math>(x-(2+i))(x-(2-i))(x-2i)(x+2i) = 0</math><br />
 
<div style="text-align:center;"><math>(x-(2+i))(x-(2-i))(x-2i)(x+2i) = 0</math><br />
 
<math>(x^2 - 4x + 5)(x^2 + 4) = 0</math><br />
 
<math>(x^2 - 4x + 5)(x^2 + 4) = 0</math><br />
<math>\displaystyle x^4 - 4x^3 + 9x^2 - 16x + 20 = 0</math></div>
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<math>x^4 - 4x^3 + 9x^2 - 16x + 20 = 0</math></div>
  
 
Thus our answer is <math>- 4 + 9 - 16 + 20 = 9\ \mathrm{(D)}</math>.  
 
Thus our answer is <math>- 4 + 9 - 16 + 20 = 9\ \mathrm{(D)}</math>.  
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
[[Category:Introductory Complex Numbers Problems]]
 

Revision as of 21:09, 30 November 2007

Solution

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, 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
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