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 $\binom{4}{1}=4$ 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 $(-4-4\sqrt{2})+(-4+4\sqrt{2})=-4-4=-8, \boxed{\text{A}}$ .

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