Difference between revisions of "2007 AMC 12A Problems/Problem 18"
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== Problem == | == Problem == | ||
| − | The | + | The polynomial <math>f(x) = x^{4} + ax^{3} + bx^{2} + cx + d</math> has real coefficients, and <math>f(2i) = f(2 + i) = 0.</math> What is <math>a + b + c + d?</math> |
<math>\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 4 \qquad \mathrm{(D)}\ 9 \qquad \mathrm{(E)}\ 16</math> | <math>\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 4 \qquad \mathrm{(D)}\ 9 \qquad \mathrm{(E)}\ 16</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | A fourth degree polynomial has four [[root]]s. Since the coefficients are real(meaning that complex roots come in conjugate pairs), the remaining two roots must be the [[complex conjugate]]s of the two given roots, | + | A fourth degree polynomial has four [[root]]s. Since the coefficients are real(meaning that complex roots come in conjugate pairs), the remaining two roots must be the [[complex conjugate]]s of the two given roots. By the factor theorem, our roots are <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 /> | ||
| Line 12: | Line 12: | ||
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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | Just like in Solution 1 we realize that the roots come in conjugate pairs. Which means the roots are <math>2i,i+2,-2i,2-i</math> | ||
| + | So our polynomial is | ||
| + | |||
| + | '''(1)''' <math>f(x) = (x-2i)(x+2i)(x-i-2)(x-2+i)</math> | ||
| + | |||
| + | Looking at the equation of the polynomial <math>f(x) = x^4+ax^3+bx^2+cx+d</math>. We see that <math>a+b+c+d = f(1)-1</math> | ||
| + | |||
| + | If we plug in <math>1</math> into equation '''(1)''' we get <math>f(1) = (1-2i)(1+2i)(-1-i)(-1+i)</math>. | ||
| + | |||
| + | Now if we multiply a complex number by its conjugate we get the sum of the squares of its real and imaginary parts. Using this property on the above we multiply and get <math>f(1) = (1-2i)(1+2i)(-1-i)(-1+i) = (1^2+2^2)(1^2+1^2) = 10</math> | ||
| + | So the answer is <math>f(1) -1 = 10 - 1 = 9</math>. <math>\framebox{D}</math> | ||
| + | |||
| + | ==Video Solution 1 by SpreadTheMatthLove== | ||
| + | https://www.youtube.com/watch?v=WgySLk2p5VU | ||
==See also== | ==See also== | ||
Latest revision as of 23:37, 24 March 2024
Problem
The polynomial 
has real coefficients, and 
What is 
Solution 1
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 
. Now we work backwards for the polynomial:
Thus our answer is 
.
Solution 2
Just like in Solution 1 we realize that the roots come in conjugate pairs. Which means the roots are 
So our polynomial is
(1) 
Looking at the equation of the polynomial 
. We see that 
If we plug in 
into equation (1) we get 
.
Now if we multiply a complex number by its conjugate we get the sum of the squares of its real and imaginary parts. Using this property on the above we multiply and get 
So the answer is 
. 
Video Solution 1 by SpreadTheMatthLove
https://www.youtube.com/watch?v=WgySLk2p5VU
See also
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
