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Let $W(x)$ be a polynomial with n real roots and $P(x) = \alpha W(x) + W'(x)$. Prove that for any $\alpha \in \mathbb{R}$: $P(x)$ have at least $n-1$ real roots.

I know that the degree of the polynomial is one higher than the degree of derivative. Yet I have no idea how to even begin.

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    $\begingroup$ Consider $f(x) = e^{\alpha x} W(x)$. Use Rolle's theorem. $\endgroup$ – peter a g Aug 4 '15 at 19:39
  • $\begingroup$ Use Rolle's theorem for $f$ from @peterag between zeros. $\endgroup$ – A.Γ. Aug 4 '15 at 19:40
  • $\begingroup$ Sorry @A.G - I re-posted at the same time as you added your comment $\endgroup$ – peter a g Aug 4 '15 at 19:41
  • $\begingroup$ @peterag It's ok, no problem ;) $\endgroup$ – A.Γ. Aug 4 '15 at 20:07
  • $\begingroup$ @peterag Why not upgrade your comment to a full answer, since it already contains the key idea? $\endgroup$ – Travis Aug 4 '15 at 20:22
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As requested... Suppose $W(x)$ is a real polynomial with $n$ real zeros. We may assume $n\ge 2$, for otherwise there is nothing to show. For the moment, assume that the roots are distinct - which, I confess (sigh, apologies), I had assumed in the above comment.

So, let $x_1 < \cdots < x_n$, be roots of $W$, i.e., $W(x_i) =0$, for $1\le i \le n$.

Let $f(x) = e^{\alpha x} W(x)$. Then $f$ vanishes precisely at the same points as $W$, i.e., $f(x_i) = 0 $; "precisely", because the exponential factor of $f$ cannot vanish.

By Rolle's theorem, there exists $y_i \in (x_i, x_{i+1})$ for $1\le i \le n-1$, such that $f'(y_i) = 0.$

Now, $$ f'(x) = \alpha e^{\alpha x} W(x) + e^{\alpha x } W'(x) = e^{\alpha x }\left(\alpha W(x) + W'(x) \right) = e^{\alpha x} P(x).$$
Substitute $y_i$ for $x$ in the previous. Then, as $f'(y_i) =0$, $$e^{\alpha y_i } P(y_i) = 0.$$ As the exponential factor in front of $P$ cannot vanish, one concludes $$P(y_i) = 0$$ for $1\le i \le n-1$.

Taking into account multiplicities - i.e., repeated roots. Suppose $$ W(x) = (x-a)^k q(x),$$ with $k\ge 2$, and $q $ a polynomial where $q(a)\ne 0$. Then $$W'(x)= (x-a)^{k-1}\,\left( k q(x) + (x-a ) q'(x)\right).$$ Therefore $P(x) $ has a factor of $ (x-a)^{k-1} $, i.e., vanishes to degree $k-1$ at $x=a$. (If $k=1$, the statement is trivially true. )

Combining the two parts of the argument: suppose $x_1 < \cdots < x_s$ are the distinct roots, with multiplicity $k_i$, so that $$ k_1 + \cdots + k_s = n.$$ Then $P(x)$ has at least $$ (k_1 - 1) + \cdots (k_s -1) + (s - 1) $$ roots. Doing the algebra, $P(x)$ has at least $n-1$ roots.

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