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

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.

• GCD of $f$ and $f'$. Oct 28, 2018 at 14:26
• yes i know that.but my question is there any easy process to find GCD of two polynomial because sometimes it is harder to factor two polynomial with degree $4$ or $5$ @Wuestenfux Sir Oct 28, 2018 at 14:29
• The division method is not particularly difficult, is it? In any case, I don't know of an easier method. Oct 28, 2018 at 14:30
• Euclidean algorithm works. Oct 28, 2018 at 14:31
• @Wuestenfux: I think that's just another name for it, isn't it? Oct 28, 2018 at 14:32

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

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.

• +1 especially for last sentence. Oct 28, 2018 at 14:50

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, $$g(x)=x+4$$.

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

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

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

• Thanks a lot @Will Jagy Sir. I appreciated your hard Working :) It let me to know new thing Oct 29, 2018 at 14:07
• Where can i learn about this technique , i just know normal euclodean algorithm for polynomials , this one went over my head . Apr 6, 2022 at 4:39
• @Orion_Pax the first part is the same as the Euclidean algorithm for $\mathbb Q[x]$ To get the Bezout expression for polynomials $f.g$ that is $fh_1 + g h_2 = 1,$ most people do "back substitution." I have come to like continued fractions, and found that they can be used to find Bezout both in the integer case and the (rational) polynomial case. Integers are easier, I suggest you learn how to solve, say, $1000x - 739 y = 1$ by finding the (finite) continued fraction for $\frac{1000}{739}$ Apr 6, 2022 at 17:30