Maths Olympiad Prep

Track / Stage 4 / 9 of 340 #269 of 1964

Problem 269

AMC 12 late, AIME early
Algebra Difficulty 4.1 Multiple choice Italian Mathematical Olympiad · Italy

How many distinct real solutions does the equation x6+2x5+2x4+2x3+2x2+2x+1=0x^{6}+2 x^{5}+2 x^{4}+2 x^{3}+2 x^{2}+2 x+1=0 have?

Pick one

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Official solution

The answer is (B). First, observe that the polynomial in the statement can be written as (1+2x+2x2+x3)+(x3+2x4+2x5+x6)=(1+2x+2x2+x3)+x3(1+2x+2x2+x3)=(1+x3)(1+2x+2x2+x3)(1+2 x+2 x^{2}+x^{3})+(x^{3}+2 x^{4}+2 x^{5}+x^{6}) = (1+2 x+2 x^{2}+x^{3}) + x^{3}(1+2 x+2 x^{2}+x^{3}) = (1+x^{3})(1+2 x+2 x^{2}+x^{3}).

Recognizing further that we can write 1+2x+2x2+x3=(1+x+x2)+(x+x2+x3)=(1+x)(1+x+x2)1+2 x+2 x^{2}+x^{3} = (1+x+x^{2}) + (x+x^{2}+x^{3}) = (1+x)(1+x+x^{2}), the problem reduces to solving the equation (x3+1)(x+1)(1+x+x2)=0(x^{3}+1)(x+1)(1+x+x^{2})=0.

Since a product is zero when one of the factors is zero, the solutions are those of the equation x3=1x^{3}=-1, namely x=1x=-1, those of x+1=0x+1=0, namely again 1-1, and those of 1+x+x2=01+x+x^{2}=0, which has no solutions, since 1+x+x2=(x+12)2+341+x+x^{2} = \left(x+\frac{1}{2}\right)^{2}+\frac{3}{4} is always positive. We conclude therefore that the answer is 1: the only real solution of the proposed equation is x=1x=-1.

Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty, ordering) added by this project.