Simple Polynomial Factorization

[naptor]

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 :

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

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


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

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

What's wrong?

 

[Office_Shredder]

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 \omega = e^{2\pi i/3}, then a\omega and a \omega^2 are also roots. So
p(x) = (x-a)(x-a\omega)(x-a\omega^2)

 

[naptor]

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 \omega = e^{2\pi i/3}, then a\omega and a \omega^2 are also roots. So
p(x) = (x-a)(x-a\omega)(x-a\omega^2)

Thank you, makes perfect sense now :) .

 

[Mentallic]

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

p(x)=(x-a)(bx^2+cx+d)

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

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

=bx^3+(c-ab)x^2+(d-ac)x-ad

And we also know that

p(x)=x^3-a^3

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

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

Thus,

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

c-ab=0

d-ac=0

-ad=-a^3, d=a^2

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

So either a=0 or a=c, but we already know a\neq 0 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 into finally get

p(x)=x^3-a^3=(x-a)(x^2+ax+a^2)

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.

 

[naptor]

Thank you Metallic.
This approach definitely looks elegant, I'll try to use it to extend to the general case x^n-a^n .

 

[Mark44]

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

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

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). )

 

[Karamata]

a is triple solution if f(a)=f'(a)=f''(a)=0.
In this case we have f(a)=a^3-a^3=0, then, f'(x)=3x^2, so f'(a)=3a^2\neq 0 (if a\neq 0), so you can't say that a is triple solution as you did (saying x^3-a^3=(x-a)(x-a)(x-a).
We supposed that a isn't function by x.

Sorry for bad English.

 

[naptor]

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\geq1) 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]

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.

 

[Mark44]

I was justifying that based on an result on algebra that, as far as I can remember, says that any n-polynomial (n\geq1) 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.

 

[naptor]

a is triple solution if f(a)=f'(a)=f''(a)=0.

So, what you're saying is that I can show that :
a is an n-ple solution \Rightarrow f(a)=f'(a)=f''(a)=...=f^n(x)=0
?

 

[Office_Shredder]

And the other way around is true also