Question

Factor each polynomial completely over the set of complex numbers.

Answer 1

Those are the directions. Then it lists a bunch of polynomials. What in the heck does that mean? I don't understand the "over the set of complex numbers" bit... do they just want me to factor the polynomial?

Answer 2

They probably want all of your factors to be linear :)
(x+something)
no x^2 allowed

Answer 3

Here is the first one they give me:

a(x) = 2x^4 -3x^3 -24x^2 + 13x + 12

Answer 4

How is (x+something) a complex number though?

Answer 5

Ahhh simmer down.
Let's see if we can get started somewhere.
Notice that all of the coefficients add up to 0.
So x=1 will be a solution to this polynomial, ya?
\(\large\rm 2-3-24+13+12=0\)

Answer 6

So we can start by applying `Synthetic Division` with x=1.
Are you familiar with that method? :)

Answer 7

Interesting, I didn't notice that. And yeah, I'll do the synthetic division and let you know what I get.

Answer 8

K cool

Answer 9

I got 2x^3 -x^2 -25x - 12

Answer 10

If all the coefficients hadn't added up to 0, how would we have found the first solution?

Answer 11

If they hadn't added up to 0,
we would be required to apply our Rational Root Theorem.
It's a little more complicated.

You take the constant term and divide it by the leading term,
12/2 = 6
RRT tells us that all rational roots must be a factor of this number.
So you can get 6 by multiplying 1 and 6, -1 and -6, 2 and 3, -2 and -3.
So those are all of the possible rational roots of this polynomial,
\(\rm \pm1,2,3,6\)

And you would check them one by one until you find a root that gives you 0.
And once you've found one, you apply Synthetic Division.

Answer 12

Ok so by applying Synthetic Division,
we've been able to turn our polynomial into
\(\rm (x-1)(2x^3-x^2-25x-12)\)

Answer 13

We pulled the x=1 root out of the rest of it.
That's what synthetic division did for us.

Answer 14

Right. Now do we do the constant divided by the leading term thing and test the numbers?

Answer 15

To factor the (2x3−x2−25x−12)

Answer 16

Mmmm ya I think so

Answer 17

Hmm 4 is a root of this polynomial.
So my explanation is missing something...
because 4 is not a factor of 6.
Mmmm sec thinking :D

Answer 18

4 is a factor of 12 though

Answer 19

Yes, I guess it was bad to simplify 12/2.

Answer 20

I would recommend checking the factors of 6 first though :)
One of those will work out.

We can do division after we find it.
And the other two ugly roots will be easily found using our `quadratic formula`.

Answer 21

Does (2x + 1)(x - 1)(x - 4)(x + 3) look about right?

Answer 22

Yes.
I dunno how you did that though :) lol

Answer 23

Lol! Just looked at the constant factors of 12 (1, 12, 2, 6, 3, 4) and divided them by the factors of the leading coefficient (2, 1). I tried 4 first using synthetic division and then just simplified the rest.

Answer 24

Anyway, thanks so much for your help! Why do you think they asked to factor each polynomial "over the set of complex numbers"? What does that even mean?

Answer 25

That's what threw me off :P

Answer 26

Ummm maybe I can give an example to show where things might have gone differently.

Let's say we were able to factor something down to: x^2+4

We could do some simply algebra to find the last two roots,
\(\large\rm x^2+4=0\)
subtract 4 from each side,
\(\large\rm x^2=-4\)
square root each side,
\(\large\rm x=\pm 2i\)
So we see that the last two roots are 2i and -2i.
So our polynomial would include factors of (x-2i) and (x+2i)

Answer 27

You wouldn't stop at x^2+4 just because it doesn't give you real roots.

Answer 28

Hopefully the square root step wasn't too confusing :U

Answer 29

No that makes sense :) Thanks so much. I have a couple more Qs... would you mind helping me out? I'll close this thread and open a new question.

Answer 30

ya open a new one :)
Ill try to hobble on over if i can spare the time :D

Answer 31

Haha! You're the best. xD