The question is:     $x^6 + 5x^3 + 8$   Please can someone help me in factorising this. I saw some solutions but they are not meant for a IX grade student.   Thanks for the help.
19792
I wanted to know, how can I factor $x^6 +5x^3 +8$, I have no idea. Is there any method to know if a polynomial is factored. Just some advice will do.  Help appreciated.  Thanks.
84140
Is there a method to find which numbers to use when simplifying quadratics?   For example $x^2 + 5x + 6$ is easy enough to find, but what if I have bigger numbers, or I have this quadratic expression: $x^2 + 7x – 6$, then it gets very hard to think of a valid simplification.  EDIT: Here's an example: $(x^2 + 5x + 6)/(x + 2)$. When simplifying the first half we get $(x + 2)(x + 3)/(x + 2)$, then we can cancel out $(x + 2)$ and the answer would be $(x + 3)$. The question is: How to always find the right numbers which to put in? (in this example they where 2 and 3)
87425
I would like to know how to factor the following polynomial.  $$ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $$ What is the method I should use to factor it? If anyone could help.. Thanks in advance.
112240
Find all ($x,y,z$) such that $$ 27^{3x^2 + 2y}+27^{3y^2 + 2z}+27^{3z^2 + 2x}=1$$  I am a high school student and would appreciate it if anyone could solve it using high school mathematics. here is what I have tried till now: $$ 3x^2 +2y &lt; 0 $$ (this can also be said for the other 3 equation. it also means that x,y,z&lt;0) $$ -2y&gt;3x^2 $$ $$ 4y^2&gt;9x^2 $$ ( -2y and 3x^2 are both positive) $$ 3y^2&gt;27/4 x^2 $$ $$ -2z&gt;3y^2&gt;27/4 x^4 $$ $$ 4z^2 &gt; 729/16 x^8 $$ $$ 3z^2&gt; 2187/64 x^8 $$ $$ -128/2187&lt;x^7 $$ $$ -2/3 &lt; x $$ $$ -2/3&lt; x,y,z &lt; 0 $$ i tried this but as you can see didn't do much good
65047
What is the coefficient of $x^{10}$ in the expansion of $(5 x^2 + 3)^{14}$?   (I would prefer to know the answer as a mathematical expression rather than a number.)  May I know how to approach this question and what will be the answer? Thanks
46482
If $a,b,c&gt;0$ and $a^3+b^3=c^3$ then prove that $$a^2+b^2-c^2&gt;6(c-a)(c-b)$$.  I tried factoring but things got more complicated. I dont think standard AM-Gm can be applied directly. Can somebody help me to proceed? Thanks a lot.
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?
54340
By viewing the polynomials as a difference of two squares, factorise the following polynomial: $$x^4+x^2+1.$$ I searched but couldn't find a way to solve this  Edit: By using Hans Lundmark hint, I get:  $$(x^2+1)^2-x^2$$   Is it fully factorized?
104468
I have seen the thread  but this didn't really have a full solution.  Is it true that if it is reducible then it can be factored into a linear factor or quadratic factor in the form $x^2 - a$. Is $a$ in the rationals? And what exactly is a linear factor.  What I did was the following:  Let $y=x^2$. Then $y^2 - 10y +1=0 \iff y^2 -10y = -1$, add $25$ to both sides:  $$y^2 - 10y + 25 = -1 + 25 \iff  (y - 5)^2 = 24 \iff y = 5 \pm 2\sqrt6$$  So $x^2=5 \pm 2 \sqrt6= (\sqrt2 \pm \sqrt3 )^2$  So $x^4 -10x^2 +1 =(x^2 -(\sqrt2 + \sqrt3 )^2)(x^2 - (\sqrt2 - \sqrt3)^2)$  Then difference of two squares:  $ (x-(\sqrt2 + \sqrt3))(x+(\sqrt2 + \sqrt3))(x-(\sqrt2 - \sqrt3))(x+(\sqrt2 - \sqrt3))=0$  Expanding this out gives $x^4 -10x^2 +1 =0$ so we have found all the roots and we can confirm that they are the roots right? None of these roots are rational so it must be irreducible?  Please can you advise me on this method that I am using and not direct me to a theorem or any type of shortcut.   Thanks in advance.
24930
It seems to be true that every real polynomial $p_n$ of degree $n$ can be factored in the following (not unique) way $$ p_n = \sum_{i=0}^n a_ix^i = s \left\{\prod_{i=1}^n(x-b_i)\right\} +t \tag{$\ast$}$$ with $a_i, s, b_i, t $ all in $\mathbb{R}$.  For example:   $x-1=(x-1)+0$ $x^2-5x+7=(x-2)(x-3)+1=(x-5)(x-0)+7$ $x^2+1 = (x-0) \cdot (x-0) +1$   I am having a hard time coming up with a rigorous proof. I have tried using the fact that every polynomial $p_n$ can be written as $(x-a)g(x)+b$, where $g$ is a polynomial function and $a$ is an arbitrary real number, but with no success. Another way it could be done is by expanding the right-hand side of $(\ast)$ and showing that the linear system of equations with the coefficients of matching powers of $x$ has always a real solution. But I feel there must be a more simple proof out there, e.g. by induction, if possible without the Fundamental Theorem of Algebra. Would be glad if someone could enlighten me!
51942
How could it possible to factorize $x^8-1$ in product of irreducibles in the rings $(\mathbb{Z}/2\mathbb{Z})[x]$ and $(\mathbb{Z}/3\mathbb{Z})[x]$?  I'm having a hard time starting the problem. Could anyone help me at this point?
64374
I found this amazingly beautiful identity . How to prove that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ without directly multiplying the factors? (I've already verified it that way). Moreover, how could someone possibly find such a factorization using complex numbers? Is it possible to find such a factorization because $A^3+B^3+C^3 - 3ABC$ is a symmetric polynomial in $A,B,C$?
24494
$ \frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\cdots $  This is a pretty unorthodox problem, and I'm not quite sure how to simplify it.  Could I get a solution?  Thanks.
91118
What will be the remainder when $f(x^{12})$ is divided by $f(x)$ where : $$f(x) = x^5 + x^4 + x^3 + x^2 + x +1$$  I have already tried but found no idea how I can do this question.
48493
I would appreciate any help with this problem: $ x^8+2x^7+2x^6+5x^5+3x^4+5x^3+2x^2+2x^1+1x^0=0 $  I know that when $x$ isn't zero $x^0=1$ so the equation could be re-written as $ x^8+2x^7+2x^6+5x^5+3x^4+5x^3+2x^2+2x+1=0 $. I am not sure what to do from here. I have tried using wolframalpha to get the solutions so I know real ones exist (2 of them, actually), but I have no idea how to get them.
44996
i would like to know how to prove that if $a^2+b^2=c^2$ then   a) $3|a$ or $3|b$  b) $5|a$ or $5|b$ or $5|c$  c) $4|a$ or $4|b$   Also that $a^2+b^2+c^2+1$ can't be divisible by 8  Thank you very much!
26381
I read that it is not always possible to solve but from Wikipedia:      Some sixth degree equations, such as $ax^6 + dx^3 + g = 0$, can be solved   by factorizing into radicals, but other sextics cannot.   So how do you solve this? I've stumbled upon this equation where; $a = n^2$, $d=m^2n-2p^2$, and $g=m^4$. Where the $a,d,g$ corresponds to the above mentioned sextic-equation and $m,n,p$ are input variables. I apologize for that ton of variables in use, might cause some confusion. Is this possible using Galois Theory or by any other means?