polynomials - $deg(P) = 4$, then $P'$ doesn't have all zeros real if $P$ has 2 complex zeros and 2 real zeros

Let $P = a_4X^4+a_3X^3+a_2X^2+a_1X+a_0 \in \mathbb{R}[$X$]$ such that $P$ has two complex zeros, $x_1 = a+bi,x_2 = a-bi, a,b \in \mathbb{R}$ and two real zeros, $x_3 = a$ and $x_4 \in \mathbb{R}$.

Prove that $P'$ doesn't have all zeros real.

I have tried writing Viete's formulas for $P$ and the reciprocal of $P$ to tie those zeros to the zeros of $P'$, but I didn't manage to solve the problem. I also thought to assume that $P'$ has all zeros real and achieve a contradiction, but I couldn't.

Answer

As explained in Olivier Begassat's comment, you can assume $P(x)=x(x^2+1)(x-c)$ for some $c\in{\mathbb R}$. Then $P(x)=x^4-cx^3+x^2-cx, P'(x)=4x^3-3cx^2+2x-c=(4x^3+2x)-c(1+3x^2)$. Since $P'$ has odd degree, it has at least one real root $\lambda$. We then have
$c=\frac{4\lambda^3+2\lambda}{1+3\lambda^2}$, and

$$P'(x)=4x^3+2x-\Bigg(\frac{4\lambda^3+2\lambda}{1+3\lambda^2}\Bigg)(1+3x^2)= (x-\lambda) \frac{Q(x)}{1+3\lambda^2}$$

where

$$Q(x)= (12\lambda^2 + 4)x^2 - 2\lambda x + (4\lambda^2 + 2)$$

The discriminant of $Q$ is $-(192\lambda^4 + 156\lambda^2 + 32) \lt 0$, so $Q$ has no real root as wished.