A pretty nice problem. Let $x_1 < x_2 < \ldots < x_n$ be the roots of $P$, then $P(x) = a(x-x_1)(x-x_2)\ldots(x - x_n).$ Now we can pair up $x_i$'s so that $P(t) < 0$ for $t \in(x_i,x_{i+1})$ (if $n$ is odd then one of $x_1$ or $x_n$ will be paired with $\pm \infty$). Now we'll prove that $\left|P\left(\frac{x_i + x_{i+1}}{2}\right)\right| > 3$, and because $P\left(\frac{x_i + x_{i+1}}{2}\right) < 0$ we would have $P\left(\frac{x_i + x_{i+1}}{2}\right) + 3 < 0$. So $$\left|P\left(\frac{x_i + x_{i+1}}{2}\right)\right| = \frac{1}{2^n} |x_i + x_{i+1} - 2x_1||x_i + x_{i+1} - 2x_2|\ldots|x_i + x_{i+1} - 2x_{i-1}||x_{i+1} - x_i||x_i - x_{i+1}||x_i + x_{i+1} - 2x_{i+2}|\ldots|x_i + x_{i+1} - 2x_n|.$$Now because $x_i$ are different, we have $|x_{i+1} - x_i| \geq1$, $|x_i+x_{i+1}-2x_{i-1}| \geq3$, $|x_i+x_{i+1}-2x_{i+2}| \geq3$ and for $k \not\in\{i-1,i,i+1,i+2\}$: $|x_i+x_{i+1}-2x_{k}| \geq3 + 4 = 7$. Therefore $$\left|P\left(\frac{x_i + x_{i+1}}{2}\right)\right| \geq \frac{1}{2^n}\cdot3^2\cdot7^{n-4} > 3,$$which is true for $n>5$. But now we a point under $-3$ for each pair of $x_i,x_{i+1}$ (if $n$ - odd one of the endpoints have $\pm \infty$), for which $P(x_i) = P(x_{i+1}) = 0$, so by IVT, $P(x) + 3$ has $n$ different real roots.