Does this polynomial have all its roots both distinct and real?

Solution 1:

What happens when you evaluate this at $x=0, 1, 1.5, 2, 2.5 \dots$? Then use the intermediate value theorem.

At the integer points the polynomial evaluates to $-1$. At alternate intermediate points the value is positive, and you get a pair of roots.

If $n$ is even you have positive values at $x=0$ and $x=n+1$ so roots at "either end" to make the full complement.

If $n$ is odd you have a positive value at $x=n+1$ which does the same.

Solution 2:

Let $Q_n(x) = P_n(x) +1 $. Chect the values at $Q_n(x)$ at points $\frac{i}{2}$ for $i = 1,3,5...,2n-1$. The signs will alternate and the value of $|Q_n(x)|$ will be greater than $1$. So, the value of $P_n(x)$ will also alternate at these points. This means that there are $n$ roots, in between these points.