Solve the equation $x^4+x^3-9x^2+11x-4=0$ which has multiple roots.

Question

Q:Solve the equation $x^4+x^3-9x^2+11x-4=0$ which has multiple roots.
My approach:Let $f(x)=x^4+x^3-9x^2+11x-4=0$.And i knew that if the equation have multiple roots then there must exist H.C.F(Highest Common Factor) of $f'(x)$ and $f(x)$ Or H.C.F of $f''(x)$ and $f(x)$.But i don't know how to find H.C.F of two polynomial by synthetic division(my book titled it and written the H.C.F).My book provided that H.C.F of $f(x),f'(x)$ and $f''(x)$ is $(x-1)$.
So $(x-1)^3$ is a factor of $f(x)$.Hence $f(x)=(x-1)^3(x+4)=0$.
Now my Question is how they find H.C.F without division method.Is there any general process to get H.C.F of polynomials without division method or some easy process.Any solution will be appreciated.
Thanks in advanced.

Answer 1

$$x^4+x^3-9x^2+11x-4=0$$

By checking the divisors of $-4$ we see that $x=1$ satisfies our equation.

Upon synthetic division, we get $$x^4+x^3-9x^2+11x-4=(x-1)^3(x+4)$$ with solutions of $$ x=1,1,1,-4$$ If you want to find the HCF of f and f', then you have to find the derivative $$ 4x^3+3x^2-18x+11$$ and find the common factors of $f$ and $f'$

Sounds like extra work for no reason.

Answer 2

There is no need to apply the Euclidean algorithm to solve this problem. Note that the sum of coefficients of $$f(x)=x^4+x^3−9x^2+11x−4$$ is $0$, so $f(1)=0$. Next, $$f'(x)=4x^3+3x^2-18x+11$$ also has zero coefficient sum. So, $f'(1)=0$. Then, $$f''(x)=12x^2+6x-18$$ also has zero coefficient sum. This gives $f''(1)=0$. Finally, $$f'''(x)=24x+6$$ does not satisfy $f'''(1)=0$. Thus, $1$ is a root of $f$ with multiplicity $3$. That is, $$f(x)=(x-1)^3g(x)$$ for some polynomial $g$. Clearly, $g$ is linear and monic, with constant term $4$ (as $(-1)^3c=-4$ implies $c=4$). So, $g(x)=x+4$.

Answer 3

I quite like the Euclidean algorithm. The Extended part can be done as a continued fraction, same steps as for integers.

$$ \left( x^{4} + x^{3} - 9 x^{2} + 11 x - 4 \right) $$

$$ \left( 4 x^{3} + 3 x^{2} - 18 x + 11 \right) $$

$$ \left( x^{4} + x^{3} - 9 x^{2} + 11 x - 4 \right) = \left( 4 x^{3} + 3 x^{2} - 18 x + 11 \right) \cdot \color{magenta}{ \left( \frac{ 4 x + 1 }{ 16 } \right) } + \left( \frac{ - 75 x^{2} + 150 x - 75 }{ 16 } \right) $$ $$ \left( 4 x^{3} + 3 x^{2} - 18 x + 11 \right) = \left( \frac{ - 75 x^{2} + 150 x - 75 }{ 16 } \right) \cdot \color{magenta}{ \left( \frac{ - 64 x - 176 }{ 75 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( \frac{ 4 x + 1 }{ 16 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 4 x + 1 }{ 16 } \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - 64 x - 176 }{ 75 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 16 x^{2} - 48 x + 64 }{ 75 } \right) }{ \left( \frac{ - 64 x - 176 }{ 75 } \right) } $$ $$ \left( x^{2} + 3 x - 4 \right) \left( \frac{ 16}{75 } \right) - \left( 4 x + 11 \right) \left( \frac{ 4 x + 1 }{ 75 } \right) = \left( -1 \right) $$ $$ \left( x^{4} + x^{3} - 9 x^{2} + 11 x - 4 \right) = \left( x^{2} + 3 x - 4 \right) \cdot \color{magenta}{ \left( x^{2} - 2 x + 1 \right) } + \left( 0 \right) $$ $$ \left( 4 x^{3} + 3 x^{2} - 18 x + 11 \right) = \left( 4 x + 11 \right) \cdot \color{magenta}{ \left( x^{2} - 2 x + 1 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x^{2} - 2 x + 1 \right) } $$ $$ \left( x^{4} + x^{3} - 9 x^{2} + 11 x - 4 \right) \left( \frac{ 16}{75 } \right) - \left( 4 x^{3} + 3 x^{2} - 18 x + 11 \right) \left( \frac{ 4 x + 1 }{ 75 } \right) = \left( - x^{2} + 2 x - 1 \right) $$

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$$ \left( 4 x^{3} + 3 x^{2} - 18 x + 11 \right) $$

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

$$ \left( 4 x^{3} + 3 x^{2} - 18 x + 11 \right) = \left( 2 x^{2} + x - 3 \right) \cdot \color{magenta}{ \left( \frac{ 4 x + 1 }{ 2 } \right) } + \left( \frac{ - 25 x + 25 }{ 2 } \right) $$ $$ \left( 2 x^{2} + x - 3 \right) = \left( \frac{ - 25 x + 25 }{ 2 } \right) \cdot \color{magenta}{ \left( \frac{ - 4 x - 6 }{ 25 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( \frac{ 4 x + 1 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 4 x + 1 }{ 2 } \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - 4 x - 6 }{ 25 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 8 x^{2} - 14 x + 22 }{ 25 } \right) }{ \left( \frac{ - 4 x - 6 }{ 25 } \right) } $$ $$ \left( 4 x^{2} + 7 x - 11 \right) \left( \frac{ 2}{25 } \right) - \left( 2 x + 3 \right) \left( \frac{ 4 x + 1 }{ 25 } \right) = \left( -1 \right) $$ $$ \left( 4 x^{3} + 3 x^{2} - 18 x + 11 \right) = \left( 4 x^{2} + 7 x - 11 \right) \cdot \color{magenta}{ \left( x - 1 \right) } + \left( 0 \right) $$ $$ \left( 2 x^{2} + x - 3 \right) = \left( 2 x + 3 \right) \cdot \color{magenta}{ \left( x - 1 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x - 1 \right) } $$ $$ \left( 4 x^{3} + 3 x^{2} - 18 x + 11 \right) \left( \frac{ 2}{25 } \right) - \left( 2 x^{2} + x - 3 \right) \left( \frac{ 4 x + 1 }{ 25 } \right) = \left( - x + 1 \right) $$