- #1
footmath
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Hello.
Galois theory tell us $x^5-6x+3$ is not solvable by radical but every equation lower than fifth can solve by radical.
If $G$ is solvable and $H$ is solvable too $G*H$ are solvable. For $x^5-6x+3$ we can use Newton’s method and find one root of this equation.
We obtain $x=1.4$ and factor this equation.
$x^5-6x+3=(x-1.4)(ax^4+px^3+qx^2+sx+t)$
$a=1$
$p=1.4$
$q=(1.4)^2$
$s=(1.4)^3$
$t=-6+(1.4)^4$
$x^5-6x+3=(x-1.4)(x^4+1.4x^3+(1.4)x^2+(1.4)^3x-6+(1.4)^4)$
$(x-1.4)$ is solvable and $(x^4+1.4x^3+(1.4)x^2+(1.4)^3x-6+(1.4)^4)$ is solvable too so $(x-1.4)*(x^4+1.4x^3+(1.4)x^2+(1.4)3x-6+(1.4)^4)$ are solvable.
Why did Galois theory say that $x^5-6x+3$ is not solvable by radical?
Galois theory tell us $x^5-6x+3$ is not solvable by radical but every equation lower than fifth can solve by radical.
If $G$ is solvable and $H$ is solvable too $G*H$ are solvable. For $x^5-6x+3$ we can use Newton’s method and find one root of this equation.
We obtain $x=1.4$ and factor this equation.
$x^5-6x+3=(x-1.4)(ax^4+px^3+qx^2+sx+t)$
$a=1$
$p=1.4$
$q=(1.4)^2$
$s=(1.4)^3$
$t=-6+(1.4)^4$
$x^5-6x+3=(x-1.4)(x^4+1.4x^3+(1.4)x^2+(1.4)^3x-6+(1.4)^4)$
$(x-1.4)$ is solvable and $(x^4+1.4x^3+(1.4)x^2+(1.4)^3x-6+(1.4)^4)$ is solvable too so $(x-1.4)*(x^4+1.4x^3+(1.4)x^2+(1.4)3x-6+(1.4)^4)$ are solvable.
Why did Galois theory say that $x^5-6x+3$ is not solvable by radical?