A polynomial of third grade with three roots, can't have one of this roots as a critical point. I was trying to solve a problem, it says like this: If $p (x)$ is a third degree polynomial and $a, b, c, d$ are real numbers with
d> a since $p (a) = p (b) = p (c) = p '(b) = p' '(c) = 0 $ and $p (d)> 0$, will there exist
some real number $x$ such that $(x - a) ⋅ p (x) +1 = 0$?
I thinked some time and I arrived that this problem is incorrect, because, if $p$ is supposed to be a third degree polynomial, and because $b$ is a root and a critical number, there's no exist such $p$.
I have this proof, I don't know if it is sufficient 'rigorous' or it is correct, thanks in advantage.
Let $p (a) = p (b) = p (c) = p '(b) = 0$ and $p$ a third degree polynomial. Because $b$ is a critical number and $p$ a third degree polynomial, can only exist other $k$ such that $p^{\prime}(k)=0$. If $b$ is a relative maximum, on the right of $b$, $p$ is decreasing, and on the left of $b$, $p$ is increasing. No matter that $b < a,c$ or  $b> a,c$ or $a<b<c$, etc. There must exist other two $q$ and $k$ such that $p'(q)=p'(k)=0$ because $p$ have other two roots, and it would need to change of increasing/decreasing two times.. But, this leads us to a contradiction, because $p$ can only have a maximum of $2$ critical points. It is similar with $b$ a relative minimum. If $b$ is not a relative maximum or a relative minimum, $p$ has to be strictly increasing or decreasing over an interval $(b-\alpha, b+\alpha)$. No matter that $b < a,c$ or  $b> a,c$ or $a<b<c$, etc. There must exist other two $q$ and $k$ such that $p'(q)=p'(k)=0$ because $p$ have other two roots, and it would need to change of increasing/decreasing two times. But, this leads us to a contradiction, because $p$ can only have a maximum of $2$ critical points. So we showed that if $p$ is a third degree polynomial with three roots, and one of these roots is a critical number, $p$ does not exist.
 A: $$p(x)=Ax^3+Bx^2+Cx+D=A(x-\alpha)(x-\beta)(x-\gamma)$$
where $\alpha,\beta,\gamma$ are the roots. Since we are given that $p(a)=p(b)=p(c)=0$ we know that there are the roots and since we only care about the sign we can get rid of $A$:
$$p(x)=(x-a)(x-b)(x-c)$$
$$p(x)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$$

we are also told that $p'(b)=0$:
$$p'(x)=3x^2-2(a+b+c)x+(ab+ac+bc)$$
$$\Rightarrow 3b^2-2(a+b+c)b+(ab+ac+bc)=0\tag{1}$$
and that $p''(c)=0$:
$$p''(x)=6x-2(a+b+c)$$
$$\Rightarrow 6c-2(a+b+c)=0$$
rearranging this we get that:
$$c=\frac{a+b}{2}\tag{2}$$
which gives us that $c$ is also the middle root. Also if we sub and rearrange $(1)$ we get:
$$3b^2-3ab-3b^2+2ab+\frac{a^2+b^2}{2}=0$$
$$\Rightarrow a^2-2ab+b^2=0$$
or:
$$(a-b)^2=0$$
which gives us that:
$$a=b=c\tag{3}$$

so we can rewrite our polynomial:
$$p(x)=A(x-a)^3$$
now we are told that $d>a$ and $p(d)>0$ so:
$$A(d-a)^3>0$$
but since $d-a>0$ that means that $A>0$.

Now finally lets look at the question:
$$\exists x\in\mathbb{R}\,\text{s.t.}\,\,\,\,(x-a)p(x)+1=0$$
well we can now rewrite this as:
$$A(x-a)^4+1=0$$
$$(x-a)^4=-\frac1A$$
now since $x,a$ are real and $A$ is positive, we can confirm such $x$ does not exist as it would not be real
A: Here is a shorter answer than the first one posted. Note first that a cubic polynomial has three roots which need not be distinct. We are given the roots $a,b,c$ so that $p(x)=A(x-a)(x-b)(x-c)$
Now the condition $p'(b)$ is equivalent to saying that there is a double root at $b$ so we must have $b=a$ or $b=c$ so we have $p(x)=A(x-a)^2(x-c)$ or $p(x)=A(x-a)(x-c)^2$
In the second case with $p''(c)=0$ we have a triple root at $c=a$ whence $p(x)=A(x-a)^3$. The condition on $d$ ensures $A\gt 0$ and then $(x-a)p(x)+1=A(x-a)^4+1$ is the sum of two non-negative terms one of which is always positive.
In the first case $p'(x)=2A(x-a)(x-c)+A(x-a)^2$ and $p''(x)=2A(x-c)+2A(x-a)+2A(x-a)$ and $p''(c)=0$ implies $4A(c-a)=0$ from which we deduce once again that $a=c$, there is a triple root and we are back to the previous case.
