Difference between revisions of "1974 AHSME Problems/Problem 3"
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<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. | ||
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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>. | 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>. | ||
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| + | ==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: | ||
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| + | - 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== | ||
Revision as of 13:06, 13 September 2025
Contents
Problem
The coefficient of 
in the polynomial expansion of
![\[(1+2x-x^2)^4\]](http://latex.artofproblemsolving.com/2/3/9/2399c6b47d2c9caaf078097a183bf782cfd3abff.png)
is
Solution 1
Let's write out the multiplication, so that it becomes easier to see.
![\[(1+2x-x^2)(1+2x-x^2)(1+2x-x^2)(1+2x-x^2)\]](http://latex.artofproblemsolving.com/5/7/7/5776793d7b7c952f044d47e440c6b5a144e6a823.png)
We can now see that the only way to get an is by taking three 
and one 
. There are 
ways to pick which term the comes from, and the coefficient of each one is 
. Therefore, the coefficient of is 
.
Solution 2
![\[(x^2-2x-1)^4=(x^2-2x+1-2)^4=((x-1)^2-(\sqrt{2})^2)^4=(x-1-\sqrt{2})^4(x-1+sqrt{2})^4\]](http://latex.artofproblemsolving.com/1/9/1/191eb36c12ff8768645f8d3d52e0e2788d522470.png)
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 
in 
is 4a, we have:
- Coefficient of (by extension ) in the first polynomial 
;
- Coefficient of (by extension ) in the second polynomial is 
.
So the coefficient of in the expansion is 
.
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. 
