# Thread: Factoring high number exponents completely

1. ## Factoring high number exponents completely

Hi, I was confused if I was factoring high number exponents completely
For example: x8-22x7+121x6
I just made it into a simple polinomial inside, by doing x6(x2-22x+121)
Then I factored it down to x6(x-11)(x-11) which also equals x6(x-11)2
Is this really factoring it down completely?

Because I did it with other polynomials such as x3+16x2+16x and turned it into x2(x+8)(x+8)
As well with 4x6-12x5+9x4 and factored it down to x4(2x-3)(2x-3)

Am I really factoring it down completely? Because I did this with all my problems, taking out the x with the lowest exponents then turning the ones into the paranthesis into simple polynomials and factoring that instead.

Thanks

2. ## Re: Factoring high number exponents completely

Well, here are a couple of questions that should help you find out. Can you factor $\displaystyle x$ or $\displaystyle x+8$ further? Can you factor $\displaystyle (ax+b)(cx+d)$ in another way?

3. ## Re: Factoring high number exponents completely

Originally Posted by Vlasev
Well, here are a couple of questions that should help you find out. Can you factor $\displaystyle x$ or $\displaystyle x+8$ further? Can you factor $\displaystyle (ax+b)(cx+d)$ in another way?
I'm not sure if I am right, but I don't think you can factor it any further because x is already simplified, as well as x + 8. Also (ax+b)(cx+d) is usually the factored form. These example problems were from my summer Calculus Homework, and it seemed too easy, so I thought I might be doing it too short or wrong.

Thanks for that! I hope that was the right answer!

4. ## Re: Factoring high number exponents completely

You are right, you cannot factor it any further because it is already linear. The polynomials are of this form $\displaystyle a$, $\displaystyle ax+b$, $\displaystyle ax^2+bx+c$, etc, where the powers of $\displaystyle x$ are non-negative integers. If you were to factor $\displaystyle x$, you'd have to factor it as two powers of $\displaystyle x$ with exponents smaller than $\displaystyle 1$, which won't work. Further, $\displaystyle (ax+b)(cx+d)$ is already in factored form when you are working over the real numbers, up to multiplying by constants. The take-home point is that once you've reached linear factors, you know for sure that you are done. Every polynomial over the real numbers can be factored into linear and quadratic terms (this is a theorem). Over the complex numbers, every polynomial can be factored into linear terms.

Now, just for interest, here is a little proof. To make this easier to prove, factor $\displaystyle (ax+b)(cx+d)$ like this

$\displaystyle ac(x+b/a)(x+d/c)$

This is essentially the same factorization.

Take $\displaystyle (x+a)(x+b)$. Suppose for a contradiction that you can factor it in another way $\displaystyle (x+c)(x+d)$. You obviously don't want to have $\displaystyle a = c$ or $\displaystyle a = d$ because this is essentially the same factorization then. Then you must have that

$\displaystyle (x+a)(x+b) = (x+c)(x+d)$

$\displaystyle x^2+(a+b)x+ab = x^2+(c+d)x+cd$

These better be the same polynomials. That is, the coefficients must match

$\displaystyle a+b = c + d, ab = cd$

From the first equation, you have $\displaystyle b = c + d - a$. Substitute this in the second equation

$\displaystyle a(c+d-a) = cd$
$\displaystyle -a^2 + (c+d)a - cd = 0$
$\displaystyle a^2 - (c+d)a +cd = 0$

Solving this gives you that $\displaystyle a = c$ or $\displaystyle a = d$. This is what happens in the real numbers. If you ever study ring theory, you will see that in some rings that are not unique factorization domains, you can actually factor some polynomials in more than one way.