So I let $P(x)=ax^5+bx^4+cx^3+dx^2+ex+f$, and obviously $f=0$. Then, $2^5a+2^4b+2^3c+2^2d+2e={1\over 2}$ $3^5a+3^4b+3^3c+3^2d+3e={2\over 3}$ $4^5a+4^4b+4^3c+4^2d+4e={3\over 4}$ $5^5a+5^4b+5^3c+5^2d+5e={4\over 5}$ and we are looking for $6^5a+6^4b+6^3c+6^2d+6e=$ Then I got stuck... Could anyone at least give me some hints or methods to continue?
121582
Fix a nonnegative integer $n$. Let $P(x)$ be a polynomial of degree $n$ (over the real numbers) such that for all $k\in\left\{0,1,\ldots,n\right\}$, we have $P(k)=\dfrac{k}{k+1}$. Find $P(n+1)$.   My attempt:  I consider a new polynomial $Q(x)=(x+1)P(x)-x$. This $Q$ is a polynomial of degree $\leq n+1$ satisfying: $Q(k)=0\quad\forall k\in\{0,1,2,\ldots,n\}$. Therefore $Q(x)=\lambda x(x-1)\cdots(x-n)$ for a certain real $\lambda$. Hence, $Q(-1)=\lambda \cdot (-1)^{n+1}(n+1)!$, but also $Q(-1)=(-1+1)P(-1)+1=1$. Comparing these, we get $\lambda \cdot (-1)^{n+1} (n+1)! = 1$, thus $\lambda=\dfrac{(-1)^{n+1}}{(n+1)!}$. Since $Q(n+1)=\lambda(n+1)!$ then $Q(n+1)=(-1)^{n+1} $, and from $P(x) = \dfrac{Q(x)+x}{x+1}$ we deduce: $$P(n+1)=\dfrac{(-1)^{n+1}+n+1}{n+2}.$$ I doubted that the answer were $P(n+1)=\frac{n+1}{n+2}$ but the term $(-1)^{n+1}$ wasn’t expected. Is there any mistake here? Does someone have an alternative proof?
3511
$$e^x − 6e^{-x} − 1 = 0$$  No idea how to solve this. If someone could show me the first one or two steps to push me in the right direction that would be great.
114632
In a recent piece of homework I needed to solve an equation of the form $ae^x + bx +c = 0$ where $a,b$ and $c$ are constants. I could not do it; no matter how I tried I either went in circles or hit a brick wall. Can someone demonstrate a general method for solving equations of this form? Please show every step in your answers.
110368
I want to factorize $x^5+ax^3+bx^2+cx+d$ if $d^2+cb^2=abd$ but I don't know how to use the second equality.I tried a lot but I cannot know how to use it for example it is $d^2$ but we have $d$ and if we calculate $d$ using that equality it would be harder.Any hints?
2904
Let $$P(x) = x^4+a_3x^3+a_2x^2+a_1x+a_0$$   $$P(x_0) = 0$$   $$P'(x_0) \not= 0$$   with $x_0$ and each $a_i$ real. Prove that $P(x)$ has a at least two real roots.   I can't figure why this is true.
91275
If :- $$a^2x^3 + b^2y^3 + c^2z^3 = p^5$$ $$ax^2 = by^2 = cz^2$$ $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{p}$$ Find $\sqrt a  + \sqrt b + \sqrt c$ in terms of $p$ .  What I Tried :- How do you use the information here? I assumed that :- $$ax^2 = by^2 = cz^2 = k$$ And then I can write the $1$st equation as :- $$k(ax + by + cz) = p^5$$ But I guess that just bring another extra variable into the scene so it does not help . How can I use the $3$rd equation? I can write it as $\frac{xyz}{xy + yz + zx} = p$ . How do you use it?
37471
We know that $f(x)=x^4+1$ is a .  My question is:     Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ has a linear factor over $\mathbb{Q}$?   Edit: Thanks for @Jyrki Lahtonen's answer, I want to do some modifications:     Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ is reducible over $\mathbb{Q}$?   Thanks in advance!
9127
Let $p(x)$ be a polynomial of degree $n$ satisfying $p(x)\geq 0$ for all $x$. That is, for all $x$, $p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \geq 0$, $a_n\neq 0$.  Show that $p(x)+p&#39;(x)+p&#39;&#39;(x)+\cdots+p^{(n)}(x)\geq 0$ for all $x$ where $p^{(i)}(x)$ is the $i^\text{th}$  derivative.  My interest: I know that, we can rewrite the LDE as follows  $$ p(x)+p&#39;(x)+p&#39;&#39;(x)+\cdots+p^{(n)}(x) = Lp(x) $$  where $L := I + D + D^2  + \cdots + D^{(n)}$. Can we say anything about a linear operator of this kind so that it does not change the sign of the input it takes? I can try to solve the question by writing for all the derivatives and factoring them using $p(x)$, but I think there should be a clever way of showing this by the properties of $L$. Can I figure out a solution by just looking at $L$ and the sign of $p(x)$ as in the question? Where should I look for that?
107875
My idea for the solution is something like this:  Since $2 | n$, $n = 2^a p_1^{e1} p_2^{e2} \cdots p_t^{et}$ where $a \geq 1$.  Then, $n/2 = \phi(2^a) \phi(p_1^{e1}) \phi(p_2^{e2}) \cdots \phi(p_t^{et})$.  Looking at the case that $t = 0$,  $n/2 = \phi(2^a) = 2^{a-1}$, and therefore $n = 2^a$.  Otherwise, $n = 2^a \phi(p_1^{e1}) \phi(p_2^{e2}) \cdots \phi(p_t^{et}) = n = 2^a p_1^{e1} p_2^{e2} \cdots p_t^{et}$. Since $\phi(n) &lt; n$, this is a contradiction, and therefore, the only $n$'s that apply are the powers of 2.  Is this right? Also, we were asked to find $n$ such that $\phi(n) = n/3$. How would I go about that one?
82829
I know that for any nonzero $x,y\in\mathbf{R}$,  $$ax^2+by^2+cxy &gt; 0,$$  where $a,b,c\in\mathbf{R}$. What can I deduce about $a$, $b$, and $c$?    For example, letting $x=1$ and $y=0$, I know that $\boxed{a&gt;0}$. Letting $x=0$ and $y=1$, I know that $\boxed{b&gt;0}$. Letting $x=y=1$, I know that $\boxed{c&gt;-(a+b)}$.  What else can I deduce about $a$, $b$, and $c$? How will I know when it's time to stop looking? (Is that last question even answerable?)
105972
Solve the following equations for $x, (x ∈ N^+)$  $$ (a)\ 2φ(x) = x;$$ $$(b)\ 3φ(x) = x;$$ $$(c)\ 4φ(x) = x.$$  Can anyone help me with this, or give me some hint?
17916
What is a polynomial $P(x)\in \mathbb{Q}[x]$ with root $\sqrt[3]{2}+\sqrt[3]{3}$?   I write $x=\sqrt[3]{2}+\sqrt[3]{3}$, so $(x-\sqrt[3]{2})^3=3$, but the expansion of the left side contains two cube roots again. What can we do?
18520
I know how to solve $ax + by = c$ using Extended Algorithm. But with more than variables, I'm lost :(.   To verify if it has an integer solution is easy, since we only need to check for $\gcd(a,b,c)|d$. Other than that, how can we find an integer solution for this equation?    Thanks, Chan
103995
Alright, I met this example solved in a book, but I want to know how did they come up with the answer:  $$ (1+q^2+q^4) = (1+q^2+q)(1+q^2-q) $$  Can you also refer me some internet source where I can learn the rules of factoring. Thank you in advance.
98791
$$ 2n^3 + 5 | 2n^4 +2n +2 - 2n^4 - 5n$$  $$= 2n^3+5 | 2-3n$$  $$3n-2≥2n^3 + 5$$  is this correct? Is there a more efficient way?
127
We have a polynomial $x^2 + 3x + 1$. There are 2 zeroes of it,  $\alpha$ and $\beta$  Now, what we need to find out is(What I couldn't) is as follows:  $$ \alpha^{2016} + \beta^{2016} + \alpha^{2014} + \beta^{2014} \over \alpha^{2015} + \beta^{2015}$$  Any ideas on how to do it?  Thanks a lot! -bone
79179
By using the substitution $p=x+\frac{1}{x}$, show that the equation $$2x^4+x^3-6x^2+x+2=0$$ reduces to $2p^2+p-10=0$.  I can't think of anything that produces a useful result, I tried writing p as $p=\frac{x^2+1}{x}$ and finding areas to substitute but have come with no progress. Could someone offer a slight hint on how to proceed?  Thanks