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"
Line 8: Line 8:
 
<math> \mathrm{(A)\ } -8 \qquad \mathrm{(B) \ }12 \qquad \mathrm{(C) \  } 6 \qquad \mathrm{(D) \  } -12 \qquad \mathrm{(E) \  }\text{none of these}  </math>
 
<math> \mathrm{(A)\ } -8 \qquad \mathrm{(B) \ }12 \qquad \mathrm{(C) \  } 6 \qquad \mathrm{(D) \  } -12 \qquad \mathrm{(E) \  }\text{none of these}  </math>
  
−
==Solution==
+
==Solution 1==
 
Let's write out the multiplication, so that it becomes easier to see.
 
Let's write out the multiplication, so that it becomes easier to see.
  
Line 14: Line 14:
  
 
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> 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>.
 +
 +
==Solution 2==
 +
 +
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>.
  
 
==See Also==
 
==See Also==

Revision as of 21:29, 14 January 2014

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 $\binom{4}{1}=4$ way 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

Note that $(1+2x-x^2)^4=((1-x)^2)^4 = (1-x)^8$. By the binomial theorem, the $x^7$ term is $\binom{8}{7} (-1)^1 (x)^7 = -8x$. Therefore the coefficient is $\boxed{\textbf{-8}}$.

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

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=1974_AHSME_Problems/Problem_3&oldid=58706"