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

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== Problem ==
 
== Problem ==
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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>
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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>
  
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==Solution==
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==Solution 1==
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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:
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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 />
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<math>x^4 - 4x^3 + 9x^2 - 16x + 20 = 0</math></div>
 
<math>x^4 - 4x^3 + 9x^2 - 16x + 20 = 0</math></div>
  
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Thus our answer is <math>- 4 + 9 - 16 + 20 = 9\ \mathrm{(D)}</math>.  
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Thus our answer is <math>- 4 + 9 - 16 + 20 = 9\ \mathrm{(D)}</math>.
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==Solution 2==
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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>
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So our polynomial is 
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'''(1)''' <math>f(x) = (x-2i)(x+2i)(x-i-2)(x-2+i)</math>
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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>
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 +
If we plug in <math>1</math> into equation '''(1)''' we get <math>f(1) = (1-2i)(1+2i)(-1-i)(-1+i)</math>.
 +
 
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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>
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So the answer is <math>f(1) -1 = 10 - 1  = 9</math>. <math>\framebox{D}</math>
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==Video Solution 1 by SpreadTheMatthLove==
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https://www.youtube.com/watch?v=WgySLk2p5VU
  
 
==See also==
 
==See also==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 23:37, 24 March 2024

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 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 $2-i,-2i$. 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 $- 4 + 9 - 16 + 20 = 9\ \mathrm{(D)}$.

Solution 2

Just like in Solution 1 we realize that the roots come in conjugate pairs. Which means the roots are $2i,i+2,-2i,2-i$ So our polynomial is

(1) $f(x) = (x-2i)(x+2i)(x-i-2)(x-2+i)$

Looking at the equation of the polynomial $f(x) = x^4+ax^3+bx^2+cx+d$. We see that $a+b+c+d = f(1)-1$

If we plug in $1$ into equation (1) we get $f(1) = (1-2i)(1+2i)(-1-i)(-1+i)$.

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 $f(1) = (1-2i)(1+2i)(-1-i)(-1+i) = (1^2+2^2)(1^2+1^2) = 10$ So the answer is $f(1) -1 = 10 - 1 = 9$. $\framebox{D}$

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

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