7. How many distinct real solutions does the equation $x^{6}+2 x^{5}+2 x^{4}+2 x^{3}+2 x^{2}+2 x+1=0$ have?
(A) 0
(B) 1
(C) 2
(D) 4
(E) 6.
7. The answer is $\mathbf{( B )}$. First, observe that the polynomial in the text can be written as $(1+2 x+$ $\left.2 x^{2}+x^{3}\right)+\left(x^{3}+2 x^{4}+2 x^{5}+x^{6}\right)=\left(1+2 x+2 x^{2}+x^{3}\right)+x^{3}\left(1+2 x+2 x^{2}+x^{3}\right)=\left(1+x^{3}\right)\left(1+2 x+2 x^{2}+x^{3}\right)$. Recognizing that it can be further written as $1+2 x+2 x^{2}+x^{3}=\left(1+x+x^{2}\right)+\left(x+x^{2}+x^{3}\right)=$ $(1+x)\left(1+x+x^{2}\right)$, the problem reduces to solving the equation $\left(x^{3}+1\right)(x+1)\left(1+x+x^{2}\right)=0$.

Since a product is zero when one of the factors is zero, the solutions are those of the equation $x^{3}=-1$, which is $x=-1$, those of $x+1=0$, which is again -1, and those of $1+x+x^{2}=0$, which has no solutions, since $1+x+x^{2}=\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}$ is always positive. Therefore, the conclusion is that the answer is 1: the only real solution to the proposed equation is $x=-1$.

(Lombardo)