1974 AHSME Problems/Problem 3

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

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 $(4)(-2)=-8, \boxed{\text{A}}$.

See Also

1974 AHSME (ProblemsAnswer KeyResources)
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