Solve [imath]x^4+2x^3+3x^2+1=0[/imath]

[Aion]

Problem.

Solve [imath]x^4+2x^3+3x^2+1=0[/imath] if we know it has a double root.


Useful theorem.

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

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


My attempt:

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

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

[math]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)[/math][math]p'(x)=(4x+34)(x^2-7x-3)+8(32x+13)=q_2(x)r_1(x)+r_2(x)[/math][math]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)[/math]
Here the degree of [imath]r_3(x)[/imath] is constant so we can't continue.

 

[Aion]

Problem.

Solve [imath]x^4+2x^3+3x^2+1=0[/imath] if we know it has a double root.


Useful theorem.

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

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


My attempt:

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

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

[math]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)[/math][math]p'(x)=(4x+34)(x^2-7x-3)+8(32x+13)=q_2(x)r_1(x)+r_2(x)[/math][math]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)[/math]
Here the degree of [imath]r_3(x)[/imath] is constant so we can't continue.

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

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

 

[Dr.Peterson]

I think you typed the problem wrong.

Which of these is correct?

Solve [imath]x^4+2x^3+3x^2+1=0[/imath] if we know it has a double root.

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

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 [imath]x^4+2x^3+3x^2+2x+1=0[/imath], not [imath]x^4+2x^3+3x^2+1=0[/imath]

 

[blamocur]

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