# Use of calculus in finding relations between roots of polynomial equations of degree greater than or equal to $2$

I recently came over a couple of questions while doing questions related to polynomial equations. One of them struck me as quite peculiar:

If four distinct points on the curve $$y=2x^4+7x^3+3x−5$$ are collinear, then find the arithmetic mean of $$X$$-coordinates of the aforesaid points.

The solution of this involves a very basic application of the sum of roots of a polynomial equation. But, it was their solution that confused me. For applying the sum of roots property, the $$X$$-coordinates of the line that cut the curve must be the roots of the equation. If so, doesn't this line then become the $$X$$-axis itself? If so, how to prove that the $$X$$-axis is the line that cuts the curve? Also, for general cases of this question where the polynomial has a degree greater than or equal to two and the coefficients are any arbitrary real numbers, does the fact hold that any line cutting this curve at $$n$$ distinct points where $$n$$ is the degree of the equation will always be the $$X$$-axis? { I believe that this question came in the Putnam but the explanation over there confused me even more}

• Please improve on the title if possibe. Commented Sep 24, 2020 at 7:57

Suppose the four points are on the quartic $$y=q(x)$$ and the line $$y=mx+c$$ then they satisfy $$p(x)=q(x)-mx-c=0$$ a quartic equation which has the same coefficients of $$x^4$$ and $$x^3$$ as $$q(x)$$. So the sum of the roots of $$p(x)$$ (Vieta relations) is the same as for $$q(x)$$.

This applies to polynomial equations of any degree above $$2$$.

• So is it correct to say that we can apply this for any polynomial p(x) with degree n , when cut by any other polynomial q(x) of degree not greater than (n-2) , at n points , then we shall get the x-coordinates of these points to be the roots of p(x)? Commented Sep 24, 2020 at 11:05

Take any $$4$$ points on the curve. They are collinear implies there is a line $$y=ax+b$$ through these $$4$$ points. Then the $$x$$-coordinates of these $$4$$ points will be the roots of the equation $$2x^4+7x^3+3x-5-ax-b=0$$. And sum of the roots of this equation will be $$-7/2$$, independent of $$a,b$$.

Also , for general cases of this question where the polynomial has a degree greater than or equal to two and the coefficients are any arbitrary real numbers , does the fact hold that any line cutting this curve at 'n' distinct points where 'n' is the degree of the equation will always be the x-axis?

No. Consider polynomial function $$W(x)$$ with degree $$n \geq2$$ that has n real roots. W(x) + x is polynomial function of same degree and is cut by the line y = x in n points, roots of polynomial W.

• How did you get that y=x will cut the curve of W(x) + x at n points, that too at the roots of W(x)? Commented Sep 24, 2020 at 9:20
• @MockingYak978 $y=W(x)+x$ For $W(x)=0$ we have $y=x$. Since $W(x)$ has $n$ roots so will $y=x$ Commented Sep 24, 2020 at 9:56
• $W(x) + x = x \equiv W(x) = 0$ i mean its depends of what you defined to be cut of curve 1. if its just common point of line and curve that it holds 2. if the curve has to actualy go "to the other side" of line you have to ensure that roots of W arent in even multiple Commented Sep 24, 2020 at 10:08