Simple Polynomial Factorization

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There is a theorem in algebra, whose name I don't recall, that states that given a polynomial and its roots I can easily factor it so for instance :

[itex]p(x)=x^2-36[/itex] ,
assuming that p(x) is a real function,

[itex]p(0)=0 \Leftrightarrow x=6,-6 [/itex]
then p(x) can be written as :
[itex]P(x)=(x-6)(x+6)[/itex]


I was trying to derive the good and old difference of two cubes expression using this factorization technique, then :

let [itex] p(x)=x^3-a^3 [/itex] and [itex]p(x)=0 \Leftrightarrow x=a[/itex] then [itex] p(x)=(x-a)(x-a)(x-a)=(x-a)(x^2 -2xa+a^2) ,[/itex] while the formula should be [itex] (x-a)(x^2+ax+a^2)[/itex]

What's wrong?
 
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Office_Shredder

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The problem is that polynomials have roots which aren't real numbers. While a is the only real root to x3-a3=0, if you let [itex] \omega = e^{2\pi i/3}[/itex], then [itex] a\omega[/itex] and [itex] a \omega^2[/itex] are also roots. So
[tex]p(x) = (x-a)(x-a\omega)(x-a\omega^2)[/tex]
 
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The problem is that polynomials have roots which aren't real numbers. While a is the only real root to x3-a3=0, if you let [itex] \omega = e^{2\pi i/3}[/itex], then [itex] a\omega[/itex] and [itex] a \omega^2[/itex] are also roots. So
[tex]p(x) = (x-a)(x-a\omega)(x-a\omega^2)[/tex]
Thank you, makes perfect sense now :) .
 

Mentallic

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Or you can even try this:

So you know that x=a is a root of p(x)=x3-a3, so we'll factorize that out and leave the other factor as a quadratic (because a linear equation multiplied by a quadratic is a cubic)

So what we have is

[tex]p(x)=(x-a)(bx^2+cx+d)[/tex]

And we want to find what the values b,c,d are. Well, we can expand!

[tex]p(x)=bx^3+cx^2+dx-abx^2-acx-ad[/tex]

[tex]=bx^3+(c-ab)x^2+(d-ac)x-ad[/tex]

And we also know that

[tex]p(x)=x^3-a^3[/tex]

So we can now equate the coefficients of each power in x.

[tex]p(x)=x^3+0x^2+0x-a^3\equiv bx^3+(c-ab)x^2+(d-ac)x-ad[/tex]

Thus,

[tex]bx^3=x^3, b=1[/tex]
(from here we'll neglect the x terms because they'll always cancel)

[tex]c-ab=0[/tex]

[tex]d-ac=0[/tex]

[tex]-ad=-a^3, d=a^2[/tex]

Substituting d=a2 back into d-ac=0 gives us
[tex]a^2-ac = a(a-c) = 0[/tex]

So either a=0 or a=c, but we already know [itex]a\neq 0[/itex] because that was our first root, and if a=0 then we're trying to factorize just x3 which is a trivial case, so a=c then. Thus we can plug all the values in to finally get

[tex]p(x)=x^3-a^3=(x-a)(x^2+ax+a^2)[/tex]

And you can use the quadratic formula to find the complex roots that Office_Shredder mentioned.

Oh and as a tip, you can quickly and easily use the method of equating the coefficients if you use some common sense such as already realizing that b=1 and d=a2 because there is only one x3 and constant term each.
 
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Thank you Metallic.
This approach definitely looks elegant, I'll try to use it to extend to the general case [tex]x^n-a^n[/tex] .
 
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I was trying to derive the good and old difference of two cubes expression using this factorization technique, then :

let [itex] p(x)=x^3-a^3 [/itex] and [itex]p(x)=0 \Leftrightarrow x=a[/itex] then [itex] p(x)=(x-a)(x-a)(x-a)=(x-a)(x^2 -2xa+a^2) ,[/itex] while the formula should be [itex] (x-a)(x^2+ax+a^2)[/itex]
I don't think this was touched on completely in the reponses following this post. What is wrong is that x3 - a3 does not factor into (x - a)3, as you show above.

Also, (x - a)3 ≠ (x - a)(x2 - 2ax + a2) as you also show above. (It is true that x3 - a3 = (x - a)(x2 - 2ax + a2). )
 
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[tex]a[/tex] is triple solution if [tex]f(a)=f'(a)=f''(a)=0[/tex].
In this case we have [tex]f(a)=a^3-a^3=0[/tex], then, [tex]f'(x)=3x^2[/tex], so [tex]f'(a)=3a^2\neq 0[/tex] (if [tex]a\neq 0[/tex]), so you can't say that [tex]a[/tex] is triple solution as you did (saying [tex]x^3-a^3=(x-a)(x-a)(x-a)[/tex].
We supposed that [tex]a[/tex] isn't function by [tex]x[/tex].

Sorry for bad English.
 
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I don't think this was touched on completely in the reponses following this post. What is wrong is that x3 - a3 does not factor into (x - a)3, as you show above.

Also, (x - a)3 ≠ (x - a)(x2 - 2ax + a2) as you also show above. (It is true that x3 - a3 = (x - a)(x2 - 2ax + a2). )
I was justifying that based on an result on algebra that, as far as I can remember, says that any n-polynomial (n[itex]\geq[/itex]1) has exactly(not sure about this "exactly") n roots. Therefore I thought that the only real root "a" found should have multiplicity 3. But I'm pretty sure now that the theorem meant n complex roots, which is much more reasonable. I could probably use some review of algebra theorems.
 

Char. Limit

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Yes indeed, the theorem means "complex" roots. Other than that, you had it correct, though. A polynomial of nth degree has n complex roots, exactly.
 
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I was justifying that based on an result on algebra that, as far as I can remember, says that any n-polynomial (n[itex]\geq[/itex]1) has exactly(not sure about this "exactly") n roots. Therefore I thought that the only real root "a" found should have multiplicity 3.
No, if a is a root of an n-th polynomial, that doesn't mean that the multiplicity has to be n. For the polynomial f(x) = x3 - a3, a is the only real root (hence of multiplicity 1).
But I'm pretty sure now that the theorem meant n complex roots, which is much more reasonable. I could probably use some review of algebra theorems.
 
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[tex]a[/tex] is triple solution if [tex]f(a)=f'(a)=f''(a)=0[/tex].
So, what you're saying is that I can show that :
a is an n-ple solution [itex]\Rightarrow[/itex] [tex]f(a)=f'(a)=f''(a)=...=f^n(x)=0[/tex]
?
 

Office_Shredder

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And the other way around is true also
 

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