Question

Suppose you divide a polynomial by a binomial. How do you know if the binomial is a factor of the polynomial? Can you show me a problem that has a binomial that is a factor of the polynomial being divided, and another problem that has a binomial that is not a factor of the polynomial being divided.

Answer 1

Somebody please help me.

Answer 2

Can't really tell until the division is done

Answer 3

What do you mean?

Answer 4

You have to divide the polynomial by the binomial first.

Answer 5

I mean in general there's no rule/trick that helps you detect if a binomial is a factor of a polynomial, you have to do the dividing work yourself to see it.

Answer 6

Are you sure?

Answer 7

Yes, in general.

Answer 8

This is part of a project I'm doing. I don't think your answer is helping.

Answer 9

What is that for?

Answer 10

Yes, I think so. But if you come up with such a rule to see if a binomial is a factor of polynomial without actually dividing it, feel free to publish your work here!

Answer 11

Can you at least help me with this: x^16 - 1/ x - 1. Simplify this by factoring and long division. @drawar

Answer 12

Well, this is a case when you have to tell immediately that x-1 is a factor of x^16 -1, and also you should write the factorization in a flash: \[=(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)\]

Answer 13

What's the next step?

Answer 14

What next step? x^16/x-1 = that stuff I wrote over there

Answer 15

But don't you have to simplify this: (x^8+1)(x^4+1)(x^2+1)(x+1) ?

Answer 16

What method did you use for this problem?

Answer 17

well, if you want, you can use complex numbers to factor out all these high degree binomials into binomials of degree 1

Answer 18

There's no magic here, just repeated uses of the famous identity a^2-b^2 = (a-b)(a+b)

Answer 19

I didn't get what you were saying in the first comment. But I'm only a little familiar with what you said in the second comment.

Answer 20

take x^2+1 for example, there's no real roots to the eq. x^2+1=0, you can factor it out using real numbers anymore. So if needed, you can make use of the complex number i, whose square equals -1, and write x^2+1 as (x+i)(x-i)

Answer 21

Hold on

Answer 22

Which is easier to use for this problem: x^16 - 1/ x - 1, factoring or long division?

Answer 23

factoring

Answer 24

Is what you showed you using factoring?

Answer 25

Yes

Answer 26

The last part which was (x + 1) shouldn't the x be x^2, because if you add the exponents that way you get x^16. When you typed x, I added up the exponents and got 15 instead of 16. (X^15 instead of x^16).

Answer 27

What I wrote is the result after simplifying the expression x^16-1/x-1 using factoring. If you want to get x^16-1, you have to multiply it by x-1, makes sense?

Answer 28

Can you show me how you did the multiplication? So I can see how you worked It out.

Answer 29

@drawar

Answer 30

just do from left to right: (x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)=(x^2-1)(x^2+1)(x^4+1)(x^8+1)=(x^4-1)(x^4+1)(x^8+1)=(x^8-1)(x^8+1)=x^16-1

Answer 31

Thanks. And this is using the factoring method, right?

Answer 32

Yes, because when you look at x^16-1, you immediately know what its factorization looks like. Long division would look messy in this case.

Answer 33

Thanks