Jutaro wrote: Let $f(x)$ be a cubic polynomial with rational coefficients. If the graph of $f(x)$ is tangent to the $x$ axis, prove that the roots of $f(x)$ are all rational. If $x=r$ is the point where the graph of $f(x)$ is tangent to the $x$ axis, then $f(r)=f'(r)=0$ and r is at least a double root. Polynomial division of $f(x)$ by $f'(x)$ gives $f(x)=(ax+b)f'(x)+px+q$, where $(a,b,p,q)$ are all rationals. So we have $pr+q=0$ and r is rational. Then polynomial division of $f(x)$ by $(x-r)^{2}$ gives $f(x)=(ux+v)(x-r)^{2}$, with $(u,v)$ rational and hence the third root is rational too. -- Patrick

actually i do not understand the solution .please help me giving a detailed explanation and solution. please help,

please someone reply

Hence all coefficients of $ f(x)$ are rationals, all coefficients of $ f'(x)$ are rationals. So pco wrote: Polynomial division of $ f(x)$ by $ f'(x)$ gives $ f(x) = (ax + b)f'(x) + px + q$, where $ (a,b,p,q)$ are all rationals. And hence $ f(r)=f'(r)=0$ , pco wrote: So we have $ pr + q = 0$ and r is rational. And then all coefficients of $ (x-r)^{2}$ are rationals, so pco wrote: Then polynomial division of $ f(x)$ by $ (x - r)^{2}$ gives $ f(x) = (ux + v)(x - r)^{2}$, with $ (u,v)$ rational and hence the third root is rational too.

I withdraw my comment.

DavyBa wrote: The statement of this problem is incorrect. If f(x) = x^3 + 2x, then its three roots are: 0, i√2, and -i√2 which are not all rational. Its graph is tangent to the X axis in x = 0. The derivative f`(0) is not equal to 0, but the second derivative f``(0) is. The statement probably should be: "all its real roots are rational". Does the polynomial f(x) = x^3 + 3x have 0 as a root of multiplicity 2? Not... right?...

No, it does not have a root of 0 multiplicity 2

DavyBa wrote: The statement of this problem is incorrect. If f(x) = x^3 + 2x, then its three roots are: 0, i√2, and -i√2 which are not all rational. Its graph is tangent to the X axis in x = 0. The derivative f`(0) is not equal to 0, but the second derivative f``(0) is. The statement probably should be: "all its real roots are rational". Actually, the graph is not tangent to the $x$-axis at $x=0$. This is called an "inflection point" or "saddle point", where it sort of curves out. These are identified when $f''(x)=0$. To solve the original problem, calculus actually is not required.

Yes, you are correct. I realized it myself just recently. The graph of this polynomial is not tangent to X axis since the first derivative at x = 0 is positive. It was a temporary blackout. Thank you for your reply. Also, thank you for posting the solution. May I ask you about one detail in this solution for my understanding? Since the multiplicity of r in such special cubic polynomial is 2, it's clear that such cubic polynomial has 3 real roots: r, r, and some real root q. It's also clear that the sum 2r+q (and the product (r^2)q, etc.) of these three real roots are rational (Vieta's formulas). How does it prove that all three roots are rational?

It's clear how to prove it by using calculus. Let's examine two types of cubic polynomials whose graphs are tangent to X axis. Firstly, consider polynomials f(x) = a(x-b)^3 with rational a and b, that have root r = b with multiplicity 3. Both first and second derivatives f`(b) = f``(b) = 0, so its graph is tangent to X axis and b is the inflection point. All three roots are r = b (which is a rational number). The second case is when root r of polynomial ax^3 + bx^2 + cx + d has multiplicity 2. Then, r is a local extremum. In this case, f(r) = f`(r) = 0. Polynomial division f(x)/f`(x) gives us the remainder px + q in which p and q are rational numbers. From f(r) = f`(r) = 0 it follows that pr + q = 0 and that r is a rational number. From this it follows by Vieta's formula that the third root is also rational. Q.E.D.