Well, here are a couple of questions that should help you find out. Can you factor or further? Can you factor in another way?
Hi, I was confused if I was factoring high number exponents completely
For example: x^{8}-22x^{7}+121x^{6}
I just made it into a simple polinomial inside, by doing x^{6}(x^{2}-22x+121)
Then I factored it down to x^{6}(x-11)(x-11) which also equals x^{6}(x-11)^{2 }Is this really factoring it down completely?
Because I did it with other polynomials such as x^{3}+16x^{2}+16x and turned it into x^{2}(x+8)(x+8)
As well with 4x^{6}-12x^{5}+9x^{4} and factored it down to x^{4}(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
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!
You are right, you cannot factor it any further because it is already linear. The polynomials are of this form , , , etc, where the powers of are non-negative integers. If you were to factor , you'd have to factor it as two powers of with exponents smaller than , which won't work. Further, 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 like this
This is essentially the same factorization.
Take . Suppose for a contradiction that you can factor it in another way . You obviously don't want to have or because this is essentially the same factorization then. Then you must have that
These better be the same polynomials. That is, the coefficients must match
From the first equation, you have . Substitute this in the second equation
Solving this gives you that or . 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.