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?

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Office_Shredder

Staff Emeritus
Gold Member
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

Homework Helper
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 in to 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

Mentor
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$\geq$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

Gold Member
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

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

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

Staff Emeritus
Gold Member
And the other way around is true also

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