Solve x4+2x3+3x2+1=0x^4+2x^3+3x^2+1=0

[Aion]

Problem.

Solve $x^4+2x^3+3x^2+1=0$ if we know it has a double root.


Useful theorem.

If $\alpha$ is a root to $p(x)$ of multiplicity $m$, then it's a root to $p'$ of multiplicity $m-1$. The multiple roots of $p$ are exactly the roots of the greatest common divisor of $p$ and $p'$.

Suppose $\alpha$ is a root to the greatest common divisor of $p$ and $p'$. Then it has multiplicity at least 1 as root for $p'$ and hence is at least a double root for $p$. If $\alpha$ is a multiple root for $p$ it is a root for $p'$ as well by the previous theorem. Let $f$ be the greatest common divisor for $p$ and $p'$, then there exist polynomials $g$ and $h$ such that $f=gp+hp'$ and we have $f(\alpha)=g(\alpha)p(\alpha)+h(\alpha)p'(\alpha)=0$ since $p(\alpha)=p'(\alpha)=0.$


My attempt:

By the above theorem the problem is reduced to finding the greatest common divisor $f$ for $p$ and $p'$ using the Euclidean algorithm. I've attempted to compute $GCD(p,p')$ but I'm not sure if I'm doing it correctly as it appears as if the remainder is less than degree 1, in which case $p$ and $p$' should not have any common roots. But I know that's not the case by the above assumption there exists some $\alpha$ such that $p(x)=(x-\alpha)^2q(x)$ for some polynomial $q(x)$ and $q(\alpha)\neq0$.Then $p'(x)=(x-\alpha)(2q(x)+(x-\alpha)q'(x))$, and clearly $\alpha$ has multiplicity 1 for $p'(x)$. Using WolframAlpha I see that $p(x)=(x^2+x+1)^2$ and $p'(x)=2(2x+1)(x^2+x+1)$. Hence $f=GCD(p,p')=x^2+x+1$. However, I'm not sure how to arrive at that conclusion myself.

Here's my attempt to use the Euclidean algorithm on $p$ and $p'$. Let $$p(x)=x^4+2x^3+3x^2+2x+1$$and $$p'(x)=4x^3+6x^2+6x+2$$
Dividing $p(x)$ by $p'(x)$ we have

$$p(x)=(\frac{x}{4}+\frac{1}{8})(4x^3+6x^2+6x+2)+-\frac{1}{4}(x^2-7x-3)=q_1(x)p'(x)+r_1(x)$$$$p'(x)=(4x+34)(x^2-7x-3)+8(32x+13)=q_2(x)r_1(x)+r_2(x)$$$$r_1(x)=(\frac{x}{32}-\frac{237}{32^2})(32x+13)+\frac{9}{32^2}=q_3(x)r_2(x)+r_3(x)$$
Here the degree of $r_3(x)$ is constant so we can't continue.

 

[Aion]

Problem.

Solve $x^4+2x^3+3x^2+1=0$ if we know it has a double root.


Useful theorem.

If $\alpha$ is a root to $p(x)$ of multiplicity $m$, then it's a root to $p'$ of multiplicity $m-1$. The multiple roots of $p$ are exactly the roots of the greatest common divisor of $p$ and $p'$.

Suppose $\alpha$ is a root to the greatest common divisor of $p$ and $p'$. Then it has multiplicity at least 1 as root for $p'$ and hence is at least a double root for $p$. If $\alpha$ is a multiple root for $p$ it is a root for $p'$ as well by the previous theorem. Let $f$ be the greatest common divisor for $p$ and $p'$, then there exist polynomials $g$ and $h$ such that $f=gp+hp'$ and we have $f(\alpha)=g(\alpha)p(\alpha)+h(\alpha)p'(\alpha)=0$ since $p(\alpha)=p'(\alpha)=0.$


My attempt:

By the above theorem the problem is reduced to finding the greatest common divisor $f$ for $p$ and $p'$ using the Euclidean algorithm. I've attempted to compute $GCD(p,p')$ but I'm not sure if I'm doing it correctly as it appears as if the remainder is less than degree 1, in which case $p$ and $p$' should not have any common roots. But I know that's not the case by the above assumption there exists some $\alpha$ such that $p(x)=(x-\alpha)^2q(x)$ for some polynomial $q(x)$ and $q(\alpha)\neq0$.Then $p'(x)=(x-\alpha)(2q(x)+(x-\alpha)q'(x))$, and clearly $\alpha$ has multiplicity 1 for $p'(x)$. Using WolframAlpha I see that $p(x)=(x^2+x+1)^2$ and $p'(x)=2(2x+1)(x^2+x+1)$. Hence $f=GCD(p,p')=x^2+x+1$. However, I'm not sure how to arrive at that conclusion myself.

Here's my attempt to use the Euclidean algorithm on $p$ and $p'$. Let $$p(x)=x^4+2x^3+3x^2+2x+1$$and $$p'(x)=4x^3+6x^2+6x+2$$
Dividing $p(x)$ by $p'(x)$ we have

$$p(x)=(\frac{x}{4}+\frac{1}{8})(4x^3+6x^2+6x+2)+-\frac{1}{4}(x^2-7x-3)=q_1(x)p'(x)+r_1(x)$$$$p'(x)=(4x+34)(x^2-7x-3)+8(32x+13)=q_2(x)r_1(x)+r_2(x)$$$$r_1(x)=(\frac{x}{32}-\frac{237}{32^2})(32x+13)+\frac{9}{32^2}=q_3(x)r_2(x)+r_3(x)$$
Here the degree of $r_3(x)$ is constant so we can't continue.

I must have been tired yesterday as I realise it's a simple calculation mistake.

$$p(x)=(\frac{x}{4}+\frac{1}{8})(4x^3+6x^2+6x+2)+\frac{3}{4}(x^2+x+1)=q_1(x)p'(x)+r_1(x)$$$$p'(x)=(4x+2)(x^2+x+1)+0=q_2(x)r_1(x)$$
Thus $f=gcd(p,p')=x^2+x+1$. Hence $f$ roots are $$\alpha=\frac{-1\pm i\sqrt{3}}{2}$$ both of multiplicity 2 to $p$. This is easily verified by dividing $p(x)$ by $x^2+x+1$.

 

[Dr.Peterson]

I think you typed the problem wrong.

Which of these is correct?

Solve $x^4+2x^3+3x^2+1=0$ if we know it has a double root.

Here's my attempt to use the Euclidean algorithm on $p$ and $p'$. Let $$p(x)=x^4+2x^3+3x^2+2x+1$$

You omitted "+2x", so my attempt to solve it failed, and my check of your work failed because I didn't notice the change in p.

 

[Aion]

I think you typed the problem wrong.

Which of these is correct?



You omitted "+2x", so my attempt to solve it failed, and my check of your work failed because I didn't notice the change in p.

Yes, sorry I noticed this error today. The problem was to solve $x^4+2x^3+3x^2+2x+1=0$, not $x^4+2x^3+3x^2+1=0$

 

[blamocur]

Hint: show that if $p$ and $p^\prime$ have common root then $q = p \mod p^\prime$ would have the same root too.