cubic polynomial equation $f(x)=0$ has a uniqe solution I would appreciate if somebody could help me with the following problem

Q: Let $f(x)=ax^3+bx^2+cx+d(a,b,c,d\in\mathbb{R},a\neq 0)$

prove that if $f(x)=f'(x)q(x)+r(r\in\mathbb{R})$,$q(x)$ a polynomial then equation $f(x)=0$ has a uniqe solution in $\mathbb{R}$
Letting $q(x)=ex+f$ where $e,f\in\mathbb R$, we have
$$ax^3+bx^2+cx+d=(3ax^2+2bx+c)(ex+f)+r.$$
Then, we get the followings :
$$a=3ae$$$$b=3af+2be$$$$c=2bf+ce$$$$d=cf+r$$
Hence, we get
$$e=\frac 13, f=\frac{4}{9a}, c=\frac{b^2}{3a}, d=\frac{b^3}{27a^2}+r.$$
Hence, since 
$$f(x)=ax^3+bx^2+\frac{b^2}{3a}x+d,$$
we have
$$f^\prime(x)=3ax^2+2bx+\frac{b^2}{3a}.$$

Since the discriminant of $f^\prime(x)=0$ is $0$, we now know that we reach what we want.
Otherwise $f$ must has only two distinct roots in $R$ of multiplicity 1 and 2 respectively. Let $x_0$ be the former and $x_1$ the latter. Then $f'(x_1)=0$, and from the equation$$f(x)=f'(x)q(x)+r$$ we find that$$r=0$$ Thus $$f(x)=f'(x)q(x)$$ Since $f'$ has order 2, then there must be another real root that is different from $x_1$of $f'$, let's denote it by $x_2$. Then $x_2$ is also a root of $f$ having multiplicity 2, a contradiction.