Does every even degree polynomial in the expansion of $\exp x$ have no real roots?  I tried the first few values and it seems like it... Is this a known result?
62942
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.  Attempts so far: Used Descartes signs stuff so possible number of real roots is $6,4,2,0$ tried differentiating the equation $4$ times and got an equation with no roots hence proving that above polynomial has $4$ real roots. But using online calculators I get zero real roots. Where am I wrong?
67599
I can't solve this problem:  Suppose $f$ is polynomial function of even degree $n$ with always $f\geq0$.  Prove that $f+f'+f''+\cdots+f^{(n)}\geq 0$.
52061
Does $f(x)=ax^b$ grow faster than $g(x)=\ln x$ for all $a, b &gt; 0$? Can I say that $f(x) &gt; g(x)$ as $x$ approaches infinity?  I thought the answer is yes, but this graph appears to be telling a different story.    Is the polynomial (the green curve) going to cross the log function (the red curve) and exceed in value for some large value of x?  If the answer is yes, does it mean that if I subtract the two functions and set it to zero, the resulting equation will have two roots? What are the roots?
104444
Question is to check :  which of the following is sufficient condition for a polynomial   $f(x)=a_0 +a_1x+a_2x^2+\dots +a_nx^n\in \mathbb{R}[x] $ to have a root in $[0,1]$.   $a_0 &lt;0$ and $a_0+a_1+a_2+\dots +a_n &gt;0$ $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots +\frac{a_n}{n+1}=0$ $\frac{a_0}{1.2}+\frac{a_1}{2.3}+\dots+\frac{a_n}{(n+1).(n+2)} =0$   First of all i tried by considering degree $1$ polynomial and then degree $2$ polynomial and then degree $3$ polnomial hoping to see some patern but could not make it out.  And then, I saw that $a_0= f(0)$ and $f(1)=a_0+a_1+a_2+\dots +a_n$.  So, if $f(0)&lt;0$ and $f(1)&gt;0$ it would be sufficient for $f$ to have root in $[0,1]$  In first case we have $a_0 &lt;0$ i.e., $f(0)&lt;0$ and $f(1)&gt;1&gt;0$.  So, first condition should be implying existence of a root in $[0,1]$  for second case, let $f(x)$ be a linear polynomial i.e., $f(x)=a_0+a_1x$  Now, $a_0+\frac{a_1}{2}=0$ implies $0\leq x=\frac{-a_0}{a_1}=\frac{1}{2}&lt; 1$ So, this might be possibly give existence in case of linear polynomials.  Now, $\frac{a_0}{1.2}+\frac{a_1}{2.3}=0$implies $0\leq x=\frac{-a_0}{a_1}=\frac{1}{3}&lt; 1$ So, this might be possibly give existence in case of linear polynomials.  So, for linear polynomials all the three conditions imply existence of a root in $[0,1]$.  But, i guess this can not be generalized for higher degree polynomial.  I think there should be some "neat idea" than checking for roots and all.  I am sure about first case but I have no idea how to consider the other two cases.   please provide some hints to proceed further.
116654
I can perform operations on polynomials. I can add, multiply, and find their roots. Despite this, I cannot define a polynomial. I wasn't in the advanced mathematics class in 8th grade, then in 9th grade I skipped the class and joined the more advanced class. This question isn't about something I don't understand; it's something I missed. My classes have not covered what really a polynomial is. I can generate one, but not define one. The internet has yielded incomplete definitions: &quot;Consisting of multiple terms&quot; or &quot;A mathematical expression containing 2 or more terms and variables.&quot; Take the following expressions for example: $2x^2-x+12-2x^2+x-12$. Consists of multiple terms, but can also be expressed as $0$. Is zero a polynomial? What about $x^{-1}$? , but I don't understand why. Is $x^2+x+1-x^{-1}-x^{-2}-x^{-3}$? a polynomial? It contains both positive and negative exponents?  tl;dr: What actually is the mathematical definition of a polynomial? Is $0$ a polynomial, and why isn't $x^{-1}$ a polynomial under this definition?
4718
Let $ P(x)=x^8 - x^7 +x^2 -x +15 $, Descartes' Rule of Signs tells us that the polynomial has 4 positive real roots , but if we group the terms as $$ P(x)= x(x-1)(x^6+1) +15 $$ we find that $ P(x) $ does not have any positive roots.  Where did I err ?
48676
Question: For every polynomial $P(x)$ of degree at n we have $$\sum_{i=0}^{n+1}(-1)^i\binom{n+1}{i}P(i)=0$$  Well I was inducting on degree of the polynomial, for $n=0$ its true and I got a bit confused in the inductive step, actually I cant figure the inductive step.   Is this property (if true) a famous one?  Thank you for your help.
93989
Let $K\subset \mathbb{C}$ be a subfield and $f\in K[t]$ an irreducible polynomial. Show that $f$ has no multiple roots in $\mathbb{C}$.   If I understand this question correctly, I must show that there is no $a \in \mathbb{C}$ such that $(t-a)^n|f$ in $F[t]$ with $n&gt;1$. So suppose $(t-a)^2|f$ and $f=(t-a)^2h$. Then we have $f'=2(t-a)h+(t-a)^2h' \Rightarrow (t-a)|f'$ so $\gcd(f,f'$) is not constant. Therefore $f$ is divisible by some square of non-constant polynomial in $F[t]$, which is a contradiction. Is my argument correct? Thank you.
93501
Problem is as stated in the title. Source is Larson's 'Problem Solving through Problems'. I've tried all kinds of factorizations with this trying to get it to the form $$k^2l^2$$ but nothing's clicking. I tried Bézout but the same expression can be written as $$a^2 + (a^2 + 1)(a+1)^2$$ which would imply that there is no real root.  Would really appreciate some help, thanks.
97382
If $p(x)$ is a polynomial of degree $n$ and $p(k) = k/(k + 1)$ for $k = 0, 1, 2, \ldots,n$, find $p(n + 1)$.  I'm not looking for a solution, I just want help starting it.    *Edit: Typo in original post,
70978
Find all polynomials $ f$ with real coefficients such that if $ n$ is a repunit, then so is $ f(n).$  [Note this is a Putnam question, so it is intended to be of easy to middling difficulty as contest problems go.]
2916
Consider the root of the polynomial $p(x) = x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_1x -1$.    Suppose that $p(x)$ has no roots in the open unit disc in a complex plane and $p(-1)=0$. Show that $p(1)= 0$.
44151
I'd like some help making this argument complete and rigorous (if it's correct - if not, help with that would be nice).  Here $k$ is a field.     Let $A_1,\ldots,A_n \subseteq k$ be infinite subsets. Then any polynomial in $k[x_1,\ldots,x_n]$ that vanishes on $A_1\times\cdots\times A_n\subseteq k^n$ must be $0$ (as a polynomial).   This is what I have ...  For the case $n=1$, a non-constant polynomial can only have as many roots as its degree, and in particular, it must have a finite number of roots. The only polynomial in one variable that has an infinite number of roots is $0$, so if a polynomial in $k[x_1,\ldots,x_n]$ vanishes on an infinite subset then it must be $0$.  For the inductive step, suppose the proposition is true for less than $n$ subsets and variables. Let $p\in k[x_1,\ldots,x_n]$ vanish on $A_1\times\cdots\times A_n$. Fix $x_n$ as some $a\in A_n$, and we have a polynomial in $n-1$ variables that vanishes on the set $A_1\times\cdots\times A_{n-1}$, so by the inductive hypothesis it must be identically $0$.  (Now it gets sketchy). Since this is true for any of the infinite values in $A_n$, and , $p$ must be $0$.
119842
Let $n$ be a positive integer, $x_0, x_1,\ldots,x_n$ be $(n+1)$ pairwise distinct real numbers, and $y_0, y_1,\ldots,y_n$ be $(n+1)$ real numbers. We have know that there exists a unique polynomial $P(x)$ where its degree is not greater than $n$.  What is the necessary and sufficient condition guaranteeing $P(x)$ has a degree of $n$.  Thank you for all comments and helping.
51046
So I was going through  my 11th class package on Quadratic equations and I saw a question to prove that a polynomial of $4$th degree with all real roots cannot have $\pm 1$ as all its coefficients.   I tried proving it using calculus, by showing that at least one consecutive maxima  and minima will lie either above or below the x axis, but couldn't solve it using that.  I also tried using Descartes Rule of Signs but couldn't solve it with that too. Any help?
46357
I gave an exercise to a student the other day, thinking I had a simple solution for it, but it seems that my solution was just bullshit (at least, not conclusive). Well, at least, I warned her I wasn't sure of my solution :-)  Here it is : $a\in[-1,1]$, show that all roots of polynomial $P=X^{n+1}-aX^n+aX-1$ have modulus $1$ (I try different ways of expressing the problem, tell me which one is better english...).  It is easy to see that    roots of $P$ are $1$, $-1$ for some special cases, and complex not real roots $z_i$ such that $\overline{z_i}$ is also a root, as $P$ is kind of reciprocal, $z$ root implies $\frac1z$ is also a root$.   All the other things I tried are not conclusive. For example, $a\in[-1,1]$ may be translated as $a=\cos\theta$ for some $\theta\in\mathbb R$, but I don't see how to use this. I tried some rewriting, but nothing seems to work. I tried to work on the modulus of a root, tried the relations roots-coefficients...  I'm quite stuck here. Could you help me please ?  Thanks.
60717
Let $f(t_1, ... , t_n)$ be a polynomial over a field $F$. Suppose there exist infinite subsets $X_1, ... , X_n$ of $F$ such that $f(x_1, ... , x_n) = 0$ for all $(x_1, ... , x_n) ∈ X_1× \cdots ×X_n$. Prove that $f$ is the zero polynomial.   Not sure how to start on this one!