How do I use the substitution $p = x + \frac{1}{x}$ to show that the equation $$2x^4+x^3-6x^2+x+2=0$$ reduces to $2p^2+p-10=0$.  Now, I can do the problem in reverse, but when I try solving $p = x + \frac{1}{x}$ for $x$ and substituting, the resulting equation is too complicated and nothing like the form required.
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
5923
In my text book, to solve cubic equations, I need to    find by trial &amp; error what $f(a)$ will make the equation 0. The factor will be $(x-a)$ then the other factor will be $Ax^2+Bx+C$ then I can solve using compare coefficient   But for te first part, is there any other way apart from trail &amp; error? Or is there some technique to guess the factor of an equation?   The question I am currently doing happens to be $2x^3+3x+4=9$   UPDATE:  $2x^3+3x+4=9$ should be $2x^3+3x+4=9x^2$   UPDATE 2  So I am going to try substituting   $x=\pm{1}, \pm\frac{1}{2}, \pm{2}, \pm{4}$  When $x=-\frac{1}{2}$, one of the provided answers,   $2(−0.5)^3−9(−0.5)^2+3(−0.5)+4=-18.75\neq 0$   did I make a mistake?
83909
Let $p$ be prime and $(\frac{-3}p)=1$, where $(\frac{-3}p)$ is Legendre symbol. Prove that $p$ is of the form $p=a^2+3b^2$.   My progress:  $(\frac{-3}p)=1 \Rightarrow$ $(\frac{-3}p)=(\frac{-1}p)(\frac{3}p)=(-1)^{\frac{p-1}2}(-1)^{\lfloor\frac{p+1}6\rfloor}=1 \Rightarrow$ $\frac{p-1}2+\lfloor\frac{p+1}6\rfloor=2k$  I'm stuck here. This is probably not the way to prove that.  Also tried this way:  $(\frac{-3}p)=1$, thus $-3\equiv x^2\pmod{p} \Rightarrow$ $p|x^2+3 \Rightarrow$ $x^2+3=p\cdot k$ stuck here too.  Any help would be appreciated.
64532
I have $P(x)$ a polynomial with degree $n$ ,$P(x) \ge 0$ for all $x \in$ real.  I have to prove that:     $f(x)=P(x)+P'(x)+P"(x)+......+P^{n}(x) \ge 0$ for all $x$.   I tried different methods to solve it but I got stuck.Any suggestion or advice is welcomed.
53764
Find all real polynomials with $P(0)=0$ and $P(x^2+1)=(P(x))^2+1$  I' like a solution for this problem  and proof verification:  My proof:      Let $a_0=0$ and $a_{n+1}=a_n^2+1$, clearly $P(a_0)=a_0$ and if $P(a_n)=a_n$ then $P(a_{n+1})=P(a_n^2+1)=(P(a_n))^2+1=a_n^2+1=a_{n+1}$. So $P(x)=x$ for infinitely many values of $x$ (because $a_n$ is strictly increasing), hence $P$ is the identity polynomial, and it clearly works.
44874
$$\text{ If } p \neq 2 \text{ is a prime, we know that: }$$ $$\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$$  According to this,prove that: $$p \equiv 1 \pmod 4 \Rightarrow x^2+1\equiv 0 \pmod p \text{ has a solution}$$ $$\text{Which is the solution?}$$ Solve: $\displaystyle{ x^2+1 \equiv 0 \pmod{29} \text{ , } x^2+1 \equiv 0 \pmod{37}}$  Inversely,prove that $x^2+1 \equiv 0 \pmod p \text{ has a solution } \Rightarrow p \equiv 1 \pmod 4$  I have thought the following:   $$\left(\left(\frac{p-1}{2}\right)!\right)^2-(-1)^{\frac{p+1}{2}} \equiv 0 \pmod p \Rightarrow \left(\left(\frac{p-1}{2}\right)!\right)^2+(-1)^{\frac{p+3}{2}} \equiv 0 \pmod p $$   As $p=4k+1$, $(-1)^{\frac{p+3}{2}}=(-1)^{\frac{4k+4}{2}}=1$  So,we have: $$\left(\left(\frac{p-1}{2}\right)!\right)^2+1 \equiv 0 \pmod p$$  So,we conclude that $x^2+1 \equiv 0 \pmod p \text{ has a solution , this one: } (\frac{p-1}{2})!$  Is it right or have I done something wrong?   So,are the solutions of $x^2+1 \equiv 0 \pmod{29} \text{ and } x^2+1 \equiv 0 \pmod{37}$,these one: $\displaystyle{(14)! \text{ and } (18)! \text{ respectively}}$ ? Are these the only solutions?? How can I show the inverse?
103153
Using the multinomial theorem, one can show that $f(n)$ is the coefficient of $x^n$ of the polynomial $(1+x+x^2)^n$. There are $3$ obvious ways to show the equation in the title:  First, you can square the $2$ sides of the equation and then multiply with $1-2x-3x^2$. In that case, the coefficients of the left formal power series seems too much.  Second, we can factorize the polynomial, $$    1-2x-3x^2=(-3)(x+1)(x-1/3)=(1+x)(1-3x)    $$ $$\mbox{and use the formula ,}\quad    \sqrt{1+F(x)}=\sum_{n\geq 0}^{}(-1)^n \frac{1}{4^n}{2n \choose    n}F(x)^n    $$ for $F(x)=x , G(x)=-3x$, then multiply the formal power series and see if the equation holds.  The third way is to use the previous method for $F(x)=-(2x+3x^2)$. In the 2 last methods the numbers arent very far from those we want, but i cant prove it. I think that i am missing some identity with binomial coefficients and thats why i cant solve it.
52143
I encountered this proving problem, I can do the proof but my question is why in the condition/premise we need $a$ to be unequal to $b$? My guess is that even $a=b$, the statement is still true, is it correct? If yes, how can we prove it? If false, are there any counter-examples?  The problem: Let $a$ and $b$ be two unequal constants. If $f(x)$ is a polynomial with integer coefficients, and $f(x)$ is divisible by $x-a$ and $x-b$, then $f(x)$ is divisible by $(x-a)(x-b)$.  My proof (just for reference): Let $q(x)$ be the quotient and $px+r$ be the remainder (where p, r are constants that we have to find) when $f(x)$ is divided by $(x-a)(x-b)$. So we have $f(x)=(x-a)(x-b)q(x)+px+r$. We then substitute $a$ and $b$ into $f(x)$ and use factor theorem, we get $pa+r$ and $pb+r$. We solve the simultaneous equations we get $p(a-b)=0$, since $a\neq b$, $p=0$ and $r=0$. So $f(x)$ is divisible by $(x-a)(x-b)$.  Helps are greatly appreciated. Thanks!
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.
121877
This function seemed to be pretty much straight forward, but my solution is incorrect. I have two questions: 1. Where did I make a mistake? 2. I learned that there are shortcuts for finding a series (substitution / multiplication / division / differentiation / integration of both sides). Is there something that I can apply here?  $$f(x)=\frac{1}{(1-x)^2}  \qquad  f(0)=1$$ $$f'(x)=\frac{2}{(1-x)^3}  \qquad  f'(0)=2$$  $$f''(x)=\frac{6}{(1-x)^4}  \qquad  f''(0)=6$$  $$f'''(x)=\frac{24}{(1-x)^5}  \qquad  f'''(0)=24$$  $$f''''(x)=\frac{120}{(1-x)^6}  \qquad  f''''(0)=120$$ The series is then $$1+\frac{2x}{2!}+\frac{6x^2}{3!}+\frac{24x^3}{4!}+\frac{120x^4}{5!}$$ and when simplified is  $$1+x+x^2+x^3+x^4$$ But the correct answer is  $$1+2x+3x^2+4x^3+5x^4 $$
80290
I have come across a power series as follows, \begin{equation} f(x)=\sum_{n=0}^\infty\frac{n^p}{n!}x^n. \end{equation} It shows that for some small $p$ values, we can derive simple closed forms:  $p=1: f(x)=e^xx$;  $p=2: f(x)=e^x(x+x^2)$;  $p=3: f(x)=e^x(x+3x^2+x^3)$;  $p=4: f(x)=e^x(x+7x^2+6x^3+x^4)$;  ...  Is it possible to derive a closed-form for this power series for general $p$?
104353
My question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.  Page 57. Exercise 12. Show that the following are polynomials by converting them to the form $\sum_{k=0}^{m}a_kx^k$ for a suitable $m$. In each case $n$ is a positive integer. $a)$ $(1+x)^{2n}.$ $b)$ $\frac{1-x^{n+1}}{1-x}, x\not=1.$ $c)$ $\prod_{k=0}^{n}(1+x^{2^k}).$  The attempt at a solution: a) part of the problem is pretty easy I guess, it is example of binomial theorem, so the answer would be $(1+x)^{2n}=\sum_{k=0}^{2n}(^{2n}_{k})x^k.$ Answer to part b) would be the following: $$\frac{1-x^{n+1}}{1-x}=\frac{(1-x)(1+x+x^2+\cdots+x^n)}{1-x}=1+x+x^2+\cdots+x^n=‌​\sum_{k=0}^{n}x^k.$$thanks to @DiegoMath's hint. As for part c), we have $$\prod_{k=0}^{n}(1+x^{2^k})=(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots(1+x^{2^n})$$ and I have trouble &quot;converting&quot; this to sum which would be of a form of polynomial.
29881
Please consider the integral:  $$\int_{-a}^{a}x^2dx=\frac{2a^3}{3}$$  I would like to know why I can't make the substitution:  $$u=x^2$$  When I make the substitution, the limits of the integral will be the same, and the integral itself will be zero, which is the wrong answer. So why does this simple change of variables not work as I have expected?  Please note that I do not want help solving the integral, I know how to solve it several ways. My question is why does this specific attempt at a solution not work?
45924
When the polynomial $P(x)$ is divided by $x^2 -1$, the remainder is $3x-1$. What is the remainder when $P(x)$ is divided by $x-1$?  I'm not sure what to do. I know that $P(1)=P(-1)=3x-1$, but I don't think that's how you're supposed to do it.  Any help would be appreciated, thanks.
8050
$$ x^2+2(1+a)x+(3a^2+4ab+4b^2+2)=0$$  I tried to make an inequality using the discriminant and I simplified it to get $$a^2+2ab+b^2+\frac{1}{2}≤0$$  But I don't know how to solve this.
53386
I have to find $P(0)$ from the polynomial with minimum degree given that  $$(x-1)^3|(P(x)+1)$$  $$(x+1)^3|(P(x)-1)$$  Plugging in $x=\pm 1$ gets something nice, also division by a polynomial of third order gives successively:  $$P(1)+1 =0; \ P'(1)=0; \ P''(1)=0$$ $$P(-1)-1 =0; \ P'(-1)=0; \ P''(-1)=0$$  Furthermore since $P(1)=-1$ and $P(-1)=1$, also $P(1)=-P(-1)$  and $$P'(1)=P''(1)=P'(-1)=P''(-1)$$ I don't see how to use this stuff.
89243
I have tried as follows. Please help to double check the proof! Thank you!  Since $x=p/q$ ($p$, $q$ are integers), $(p/q)^2+(p/q)a+b=0$  So, $(p/q)^2=-b-a(p/q)$  then, $p^2=-bq^2-a(p/q)q^2$  and, $p = \dfrac {q(-bq-a(p/q)q)}{p}$  now, it is clear that $q \mid p$, thus $p/q$ must be an integer.