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Factoring $2 + \frac{4}{2Z + 1} - \frac{3}{Z} + \frac{Z}{2Z^2 - Z}$

You almost got it $$\frac{8Z^3 - 2Z^2 - 5Z + 3}{Z(4Z^2 - 1)}=\frac{8Z^3 - 2Z^2 - 5Z + 3}{Z(2Z-1)(2Z+1)}$$ I am pretty sure this is the answer you were supposed to give , although this whole fraction ...
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Help factoring this third degree polynomial.

We know that $P_3(x)=ax^3+bx^2+cx+d=a(x-r_1)(x-r_2)(x-r_3)$ where $r_1,r_2,r_3$ are roots of $P_3$. (I couldn't find a neat way of solving it so here is my approach) $$11x^3+16x^2-7x-8=0$$ First we ...
ράτ's user avatar
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Help factoring this third degree polynomial.

Factoring solves the equation: $$ 11x^3+16x^2-7x-8=0\implies 11(x-x_1)(x-x_2)(x-x_3)=0$$ Which have the solutions (as found here): $$ x_1= \frac1{66} \left(-32 - \frac{(487 (1 + i \sqrt{3}))}{\sqrt[3]{...
Masd's user avatar
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Does it make sense to say a quintic equation has a "repeated quadratic root"?

Side note, we can check for a repeated factor by finding the GCD of the original polynomial $f(x)$ and its (formal) derivative $f'(x).$ In this case all coefficients for $f'$ were even, ...
Will Jagy's user avatar
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Roots of rational equation with multiple variables?

There is quite pretty geometric proof that generic quadric polynomial over algebraically closed field (this work more generally for any degree $d>1$ and number of variables $n > 2$, but I will ...
Alexander Golys's user avatar
1 vote

For $k\subset F \subset E$ algebraic field extensions, if "all" irreducible polynomials with a root in E factor in $F[x]$ then $F=E$?

Another example: consider $k=\mathbb{Q}$, $F=\mathbb{Q}[\sqrt{2}]$, $E=\mathbb{Q}[\sqrt[4]{2}]$. The only field $L$ with $k\subseteq L\subseteq E$ and $[L:k]=2$ is $F$. This follows from Galois Theory:...
Arturo Magidin's user avatar
1 vote
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For $k\subset F \subset E$ algebraic field extensions, if "all" irreducible polynomials with a root in E factor in $F[x]$ then $F=E$?

$k=\mathbb{F}_p$, $F=\mathbb{F}_{p^2}$, $E=\mathbb{F}_{p^4}$ is a counterexample. Let $f \in k[x]$ be the minimal polynomial of $a \in F \setminus E$. Now $deg(f)$ divides 4, but if $deg(f) \leq 2$ ...
Alex Day's user avatar
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Other methods of solving this question: finding $2y^4-8y^3-5y^2+26y-28$ for $y=1+\sqrt2+\sqrt3$

I would like here to stress the help that a software like SAGE can bring, using the concept of minimal polynomial. Here is the program : ...
Jean Marie's user avatar
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Other methods of solving this question: finding $2y^4-8y^3-5y^2+26y-28$ for $y=1+\sqrt2+\sqrt3$

$$\begin{align*} y-1-\sqrt 2=\sqrt 3 \\ \implies y^2+1+2-2y-2\sqrt 2y+2\sqrt 2=3\\ \implies y^2-2y=2\sqrt2(y-1)=2\sqrt 2(\sqrt 2+\sqrt 3) \\ \implies y^2-2y=4+2\sqrt 6\end{align*}$$ Input this value ...
Gwen's user avatar
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2 votes
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Other methods of solving this question: finding $2y^4-8y^3-5y^2+26y-28$ for $y=1+\sqrt2+\sqrt3$

$$ y = 1 + \sqrt{2} + \sqrt{3} \Longrightarrow (y - 1)^2 = (\sqrt{2} + \sqrt{3})^2 = 5 + 2 \sqrt{6} $$ Bring $5$ to the LHS, and then square both side again, $$ (y - 1)^2 - 5 = 2\sqrt{6} \...
Thành Nguyễn's user avatar

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