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 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 $\boxed{\textbf{-8}}$ .

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1974 AHSME Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See Also