How to find out if a polynomial has multiple roots?

Question

Find out if polynomials have multiple roots:

1, $x^6-6x^4-4x^3+9x^2+12x+4 \in {\displaystyle \mathbb {Q}}$[$x$]

2, $x^7+x^4+x^3+x+2\in {\displaystyle \mathbb {Z_3}}$[$x$]

I tried to differentiate these polynomials, but I haven't had any results. I know if a polynomial has double roots, then the first derivative of these polynomials has this root as a simple, etc.

Thank you for any help!

Answer 1

just the first one. The second polynomial is the derivative of the first but divided through by $6.$ This is the extended Euclidean algorithm, where I write the back-substitution phase in the style of a continued fraction.

$$ \left( x^{6} - 6 x^{4} - 4 x^{3} + 9 x^{2} + 12 x + 4 \right) $$

$$ \left( x^{5} - 4 x^{3} - 2 x^{2} + 3 x + 2 \right) $$

$$ \left( x^{6} - 6 x^{4} - 4 x^{3} + 9 x^{2} + 12 x + 4 \right) = \left( x^{5} - 4 x^{3} - 2 x^{2} + 3 x + 2 \right) \cdot \color{magenta}{ \left( x \right) } + \left( - 2 x^{4} - 2 x^{3} + 6 x^{2} + 10 x + 4 \right) $$ $$ \left( x^{5} - 4 x^{3} - 2 x^{2} + 3 x + 2 \right) = \left( - 2 x^{4} - 2 x^{3} + 6 x^{2} + 10 x + 4 \right) \cdot \color{magenta}{ \left( \frac{ - x + 1 }{ 2 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( x \right) } \Longrightarrow \Longrightarrow \frac{ \left( x \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - x + 1 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - x^{2} + x + 2 }{ 2 } \right) }{ \left( \frac{ - x + 1 }{ 2 } \right) } $$ $$ \left( x^{2} - x - 2 \right) \left( \frac{ 1}{2 } \right) - \left( x - 1 \right) \left( \frac{ x }{ 2 } \right) = \left( -1 \right) $$ $$ \left( x^{6} - 6 x^{4} - 4 x^{3} + 9 x^{2} + 12 x + 4 \right) = \left( x^{2} - x - 2 \right) \cdot \color{magenta}{ \left( x^{4} + x^{3} - 3 x^{2} - 5 x - 2 \right) } + \left( 0 \right) $$ $$ \left( x^{5} - 4 x^{3} - 2 x^{2} + 3 x + 2 \right) = \left( x - 1 \right) \cdot \color{magenta}{ \left( x^{4} + x^{3} - 3 x^{2} - 5 x - 2 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x^{4} + x^{3} - 3 x^{2} - 5 x - 2 \right) } $$ $$ \left( x^{6} - 6 x^{4} - 4 x^{3} + 9 x^{2} + 12 x + 4 \right) \left( \frac{ 1}{2 } \right) - \left( x^{5} - 4 x^{3} - 2 x^{2} + 3 x + 2 \right) \left( \frac{ x }{ 2 } \right) = \left( - x^{4} - x^{3} + 3 x^{2} + 5 x + 2 \right) $$

Answer 2

Hint: your first polynomial can be written as $$\left( x-2 \right) ^{2} \left( x+1 \right) ^{4}$$ and your second polynomial as $$\left( x+1 \right) \left( {x}^{6}-{x}^{5}+{x}^{4}+{x}^{2}-x+2 \right) $$