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).
Recognizing further that we can write 1+2x+2x2+x3=(1+x+x2)+(x+x2+x3)=(1+x)(1+x+x2), the problem reduces to solving the equation (x3+1)(x+1)(1+x+x2)=0.
Since a product is zero when one of the factors is zero, the solutions are those of the equation x3=−1, namely x=−1, those of x+1=0, namely again −1, and those of 1+x+x2=0, which has no solutions, since 1+x+x2=(x+21)2+43 is always positive. We conclude therefore that the answer is 1: the only real solution of the proposed equation is x=−1.