# 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 $$a, etc. There must exist other two $$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, 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.

• Do you know that if $p(b)=p'(b)=0$ for a polynomial $p(x)$ then $b$ is a double root - ie $p(x)=(x-b)^2q(x)$ for some polynomial $q(x)$? To prove this consider $p(x)=(x-b)s(x)$ so that $p'(x)=(x-b)s'(x)+s(x)$ and $p'(b)=s(b)=0$. – Mark Bennet May 8 at 12:20

$$p(x)=Ax^3+Bx^2+Cx+D=A(x-\alpha)(x-\beta)(x-\gamma)$$ 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

• This is quite a large answer so I apologize in advance if I have missed any mistakes in my checking – Henry Lee May 8 at 12:20

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.