Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 1)
Line 5: Line 5:
  
 
==Solution 1==
 
==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, namely <math>2-i,-2i</math>. Now we work backwards for the polynomial:
+
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 />

Revision as of 01:41, 19 May 2020

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 $f(1) -1 = 10 - 1 = 9$. $\framebox{D}$

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_18&oldid=122637"