If the polynomials $f(x)$ and $g(x)$ are written on a blackboard then we can also write down the polynomials $f(x)\pm g(x)$, $f(x)g(x)$, $f(g(x))$ and $cf(x)$, where $c$ is an arbitrary real constant. The polynomials $x^3-3x^2+5$ and $x^2-4x$ are written on the blackboard. Can we write a nonzero polynomial of form $x^n-1$ after a finite number of steps?
Problem
Source: All Russian 2014 Grade 11 Day 2 P3
Tags: algebra, polynomial, calculus, derivative, function, algebra proposed
leader
01.05.2014 01:55
My solution(it's the same as the official one). Nope we can't. $2$ will always be a root of the derrivatives of $F(x)$ for any $F(x)$ on the board and $2$ is not a root of $nx^{n-1}$ for any integer $n\ge 1$.
sabbasi
26.06.2014 16:59
What's the motivation behind taking the derivative of the functions? Or subbing in x=t+2 like the second official solution?
va2010
24.01.2016 22:47
I failed to solve this problem, but something about this problem is worth noting. Observe that if we let $f(x) = x^3 - 3x^2 + 5$ and $g(x) = x^2-4x$, if we can find two values $a$ and $b$ such that $f(a) = f(b)$ and $g(a) = g(b)$, then we will be done, because $f(a) = f(b)$ for all polynomials which can be created from these relations. However, if you try to solve this, you get a polynomial with a triple root at $a = b = 2$, which might inspire you to take a derivative.
Tilak_Mani
28.09.2016 19:52
Can someone please provide the official solution. A link will do. THanks.
AwesomeYRY
16.09.2021 01:26
What an amazing problem. I love russians.
Note that $P_1(x)' = 3x^2-6x = 3x(x-2)$ and $P_2(x)' = 2(x-2)$. Thus, initially all polynomial written on the board have derivative 0 at $x=2$. Note that this is invariant under the new polynomials, clearly $f(x)\pm g(x)$ work, $\frac{d}{dx} f(x)g(x) = g(x)*f'(x) + g'(x)f(x)$ works, $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$ works and also clearly $cf(x)$ works. Thus, all polynomials that are written will have derivative 0 at $x=2$.
Thus, the answer is no because the derivative of $x^n-1$ is $nx^{n-1}$ which will clearly never be negative at $x=2$ for $n>0$. And the $n=0$ case is dealt with because the question states nonzero polynomials. So we're done! The answer is no. $\blacksquare$.
HamstPan38825
04.02.2024 00:07
Notice that the derivatives $f'(x)$ and $g'(x)$ of the two existing polynomials both have zeroes at $2$. So any of $f(x) \pm g(x), f(x)g(x), f(g(x))$, and $cf(x)$ will have their derivative equal zero at $2$ by chain/product rule. But $x^n-1$ does not have a horizontal tangent at $2$, so the answer is no.
chakrabortyahan
21.03.2024 19:20
Too cute!!!!! Note that the derivatives of the initial polynomials have a common root $2$ and note that if $f'(x)$ and $g'(x) $ have common root ( whih is $2$ here) then derivatives of all the given polynomials will have that root in common as $nx^{n-1}$ does not have $2$ as a root so it is not achievable.
GMMeowChand
27.11.2024 11:55
Can't it be solved by using mod 5? But c.f(x) can be disturbing but can we generalize a bit more so that we can use mod 5?