Simple Polynomial Factorization

So, let's say that a is a double solution. This means that :f(a)=f'(a)=0 \Rightarrow f(x)=(x-a)^2*g(x) where f(x) is already factorized. If I differentiate this: f'(x)=2(x-a)g(x)+(x-a)^2*g'(x)=0 \Rightarrow 2(x-a)g(x)=-(x-a)^2*g'(x)=0 , so we have g(x)=0 and g'(x)=0 if x=a. So we have p(0)=p'(0)=0 \Leftrightarrow x=0 , but this means that x=0 is a root of multiplicity 3 of
  • #1
naptor
13
0
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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  • #2
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]
 
  • #3
Office_Shredder said:
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 :) .
 
  • #4
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 into 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.
 
  • #5
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] .
 
  • #6
naptor said:
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). )
 
  • #7
[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.
 
  • #8
Mark44 said:
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.
 
  • #9
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.
 
  • #10
naptor said:
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).
naptor said:
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.
 
  • #11
Karamata said:
[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]
?
 
  • #12
And the other way around is true also
 

1. What is simple polynomial factorization?

Simple polynomial factorization is the process of breaking down a polynomial expression into simpler, smaller expressions. It involves finding the factors of the polynomial and writing it as a product of those factors.

2. Why is simple polynomial factorization important?

Simple polynomial factorization is important because it helps us solve equations, simplify expressions, and find roots of polynomial functions. It also plays a crucial role in calculus and other branches of mathematics.

3. How do you factor a simple polynomial?

The first step in factoring a simple polynomial is to look for common factors among all the terms. Then, we use methods such as grouping, difference of squares, and trinomial factoring to further simplify the expression. Finally, we can use the quadratic formula to factor any remaining quadratic terms.

4. What are the common mistakes made in simple polynomial factorization?

Some common mistakes made in simple polynomial factorization include forgetting to factor out a common factor, making errors in the sign of the factors, and incorrectly applying the factoring methods. It is important to carefully check the final factored expression to ensure it is correct.

5. Can all polynomials be factored using simple methods?

No, not all polynomials can be factored using simple methods. Some polynomials, such as prime polynomials and irreducible polynomials, cannot be factored any further. In these cases, we can use more advanced techniques such as the rational root theorem or the quadratic formula to find the roots of the polynomial.

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