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 "1974 AHSME Problems/Problem 3"
m (→‎Solution 2)
 
(3 intermediate revisions by 3 users not shown)
Line 13: Line 13:
 
<cmath> (1+2x-x^2)(1+2x-x^2)(1+2x-x^2)(1+2x-x^2) </cmath>
 
<cmath> (1+2x-x^2)(1+2x-x^2)(1+2x-x^2)(1+2x-x^2) </cmath>
  
−
We can now see that the only way to get an <math> x^7 </math> is by taking three <math> -x^2 </math> and one <math> 2x </math>. There are <math> \binom{4}{1}=4 </math> way to pick which term the <math> 2x </math> comes from, and the coefficient of each one is <math> (-1)^3(2)=-2 </math>. Therefore, the coefficient of <math> x^7 </math> is <math> (4)(-2)=-8, \boxed{\text{A}} </math>.
+
We can now see that the only way to get an <math> x^7 </math> is by taking three <math> -x^2 </math> and one <math> 2x </math>. There are <math> \binom{4}{1}=4 </math> ways to pick which term the <math> 2x </math> comes from, and the coefficient of each one is <math> (-1)^3(2)=-2 </math>. Therefore, the coefficient of <math> x^7 </math> is <math> (4)(-2)=-8, \boxed{\text{A}} </math>.
  
 
==Solution 2==
 
==Solution 2==
 +
<cmath> (x^2-2x-1)^4=(x^2-2x+1-2)^4=((x-1)^2-(\sqrt{2})^2)^4=(x-1-\sqrt{2})^4(x-1+\sqrt{2})^4 </cmath>
 +
Each of the two polynomials can have terms of order 4,3,2,1,0. The only way to obtain a term of order 7 is either a term of order 4 from the first and 3 from the second, or the other way around. Notice that the terms of order 4 must necessarily have coefficient 1. Also keeping in mind that the coefficient of <math>x^3</math> in <math>(x+a)^4</math> is 4a, we have:
  
−
Note that <math>(1+2x-x^2)^4=((1-x)^2)^4 = (1-x)^8</math>. By the binomial theorem, the <math>x^7</math> term is <math>\binom{8}{7} (-1)^1 (x)^7 = -8x</math>. Therefore the coefficient is <math>\boxed{\textbf{-8}}</math>.
+
- Coefficient of <math>x^3</math> (by extension <math>x^7</math>) in the first polynomial <math>=4\times (-1-\sqrt{2})</math>;
 +
 
 +
- Coefficient of <math>x^3</math> (by extension <math>x^7</math>) in the second polynomial is <math>4\times (-1+\sqrt{2})</math>.
 +
 
 +
So the coefficient of <math>x^7</math> in the expansion is <math>(-4-4\sqrt{2})+(-4+4\sqrt{2})=-4-4=-8, \boxed{\text{A}}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 13:07, 13 September 2025

Problem

The coefficient of $x^7$ in the polynomial expansion of

\[(1+2x-x^2)^4\]

is

$\mathrm{(A)\ } -8 \qquad \mathrm{(B) \ }12 \qquad \mathrm{(C) \ } 6 \qquad \mathrm{(D) \ } -12 \qquad \mathrm{(E) \ }\text{none of these}$

Solution 1

Let's write out the multiplication, so that it becomes easier to see.

\[(1+2x-x^2)(1+2x-x^2)(1+2x-x^2)(1+2x-x^2)\]

We can now see that the only way to get an $x^7$ is by taking three $-x^2$ and one $2x$. There are $\binom{4}{1}=4$ ways to pick which term the $2x$ comes from, and the coefficient of each one is $(-1)^3(2)=-2$. Therefore, the coefficient of $x^7$ is $(4)(-2)=-8, \boxed{\text{A}}$.

Solution 2

\[(x^2-2x-1)^4=(x^2-2x+1-2)^4=((x-1)^2-(\sqrt{2})^2)^4=(x-1-\sqrt{2})^4(x-1+\sqrt{2})^4\] Each of the two polynomials can have terms of order 4,3,2,1,0. The only way to obtain a term of order 7 is either a term of order 4 from the first and 3 from the second, or the other way around. Notice that the terms of order 4 must necessarily have coefficient 1. Also keeping in mind that the coefficient of $x^3$ in $(x+a)^4$ is 4a, we have:

- Coefficient of $x^3$ (by extension $x^7$) in the first polynomial $=4\times (-1-\sqrt{2})$;

- Coefficient of $x^3$ (by extension $x^7$) in the second polynomial is $4\times (-1+\sqrt{2})$.

So the coefficient of $x^7$ in the expansion is $(-4-4\sqrt{2})+(-4+4\sqrt{2})=-4-4=-8, \boxed{\text{A}}$.

See Also

1974 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1974_AHSME_Problems/Problem_3&oldid=260539"