# Homework Help: Factoring Polynomials [Abstract Algebra]

1. Feb 20, 2016

### RJLiberator

1. The problem statement, all variables and given/known data

1. Let g(x) = x^4+46.
a) Factor g(x) completely in ℚ[x].
b) Factor g(x) completely in ℝ[x].
c) Factor g(x) completely in ℂ[x].

2. Completely factor the given polynomial in ℤ_5.
[4]_5 x^3 + [2]_5 x^2 + x + [3]_5

2. Relevant equations
ℚ = {m/n / m and n belong to Z, m is not divided by m}
ℝ = {real numbers which include all rational and irrational numbers}
ℂ = Complex

A polynomial is completely factored over F if it is written as a product of monic irreducible polynomials in F[x] and an element of F.

3. The attempt at a solution

Initial thoughts: This looks easy! However, it's a bit tougher than it looked.

We have x^4+46.
To start, let's just factor it as I normally would without worry about the fields.

With a little help from wolfram alpha, and a personal check to make sure we see that for the factorization over ℂ (part c) we get:
(x+4throot(-46)) (x-4throot(-46)) (x-i*4throot(-46)) (x+i*4throot(-46))

This should be the correct answer to c as it is in completely factored form.

Now, we try to get the factorization over the reals, ℝ. part b:
Since we cant use i, we try to multiply the factors a bit to get rid of i.
We get : (x^2-4throot(-46)) (x^2+4throot(-46))

Question: Is this ok over field ℝ? I have a negative under the square root, but I'm not using imaginary unit i.
Answer: Likely not as this is an expression of a+ib.
So, this doesn't work.

At this point, I'm kind of lost. I feel like, over the field ℚ, the polynomial g(x) is completely factored.
But I see to have come to the conclusion that over the field ℝ it is also completely factored.

1.a It is already completely factored
1.b It is completely factored
1.c I factored it above.

As for part 2, the difficulty here is dealing with the mod 5.
If we look at it like 4x^3+2x^2+x+3
I would think it would be in the form:
(x+)(x+)(x+)

But I'm not really sure on how to start this one. I understand what the mod 5 does, but I'm not used to trying to factor x^3+x^2+x+a.

2. Feb 20, 2016

### Ray Vickson

Try writing $p(x) = 4 x^3 + 2 x^2 + x+3$ as either $p(x) = (4 x^2+ ax + b)(x+c)$ or as $p(x) = (x^2 + ax + b)(4x+c)$, then solving for the simplest combinations of $(a,b,c)$ that work in each case.

3. Feb 20, 2016

### Staff: Mentor

It's not clear to me what your notation above means.
These look OK to me. Notationwise, you can write $\sqrt[4]{-46}$ as \sqrt[4]{-46}. The fourth roots of -46 will be arranged equidistant around a circle: at $\pi/4, 3\pi/4, 5\pi/4, 7\pi/4$.
Right. $x^4 + 46$ can't be factored over the reals.
, post: 5385224, member: 504241"]1. The problem statement, all variables and given/known data

1. Let g(x) = x^4+46.
a) Factor g(x) completely in ℚ[x].
b) Factor g(x) completely in ℝ[x].
c) Factor g(x) completely in ℂ[x].

2. Completely factor the given polynomial in ℤ_5.
[4]_5 x^3 + [2]_5 x^2 + x + [3]_5

2. Relevant equations
ℚ = {m/n / m and n belong to Z, m is not divided by m}
ℝ = {real numbers which include all rational and irrational numbers}
ℂ = Complex

A polynomial is completely factored over F if it is written as a product of monic irreducible polynomials in F[x] and an element of F.

3. The attempt at a solution

Initial thoughts: This looks easy! However, it's a bit tougher than it looked.

We have x^4+46.
To start, let's just factor it as I normally would without worry about the fields.

With a little help from wolfram alpha, and a personal check to make sure we see that for the factorization over ℂ (part c) we get:
(x+4throot(-46)) (x-4throot(-46)) (x-i*4throot(-46)) (x+i*4throot(-46))

This should be the correct answer to c as it is in completely factored form.

Now, we try to get the factorization over the reals, ℝ. part b:
Since we cant use i, we try to multiply the factors a bit to get rid of i.
We get : (x^2-4throot(-46)) (x^2+4throot(-46))

Question: Is this ok over field ℝ? I have a negative under the square root, but I'm not using imaginary unit i.
Answer: Likely not as this is an expression of a+ib.
So, this doesn't work.

At this point, I'm kind of lost. I feel like, over the field ℚ, the polynomial g(x) is completely factored.
But I see to have come to the conclusion that over the field ℝ it is also completely factored.

1.a It is already completely factored
1.b It is completely factored
1.c I factored it above.

As for part 2, the difficulty here is dealing with the mod 5.
If we look at it like 4x^3+2x^2+x+3
I would think it would be in the form:
(x+)(x+)(x+)

But I'm not really sure on how to start this one. I understand what the mod 5 does, but I'm not used to trying to factor x^3+x^2+x+a.[/QUOTE]

4. Feb 20, 2016

### RJLiberator

Ah, yes, how could I forget such things!

Marvellous.

We have:
p(x) = (4x^2+(-2)x+3)(x+1)

Which seems to be the most factorable possibility. To see if it works in mod[5] we multiply it out.

And it does.

However, I can't help but feel that this almost seems too easy here. Perhaps I need to factor it to:
x(x(4x+2)+1)+3

5. Feb 20, 2016

### RJLiberator

So my thoughts were confirmed. Part 1 cannot be factored over Q or R. Is there any way I should show this?

6. Feb 20, 2016

### Samy_A

$\mathbb Z_5$ isn't that big, you could try to find all the roots.

7. Feb 20, 2016

### Staff: Mentor

Sure. Suppose that $x^4 + 46 = 0$. Clearly x = 0 is not a solution. If x > 0, there are clearly no solutions, as both $x^4$ and 46 are positive. The same is true if x < 0. This means there are no solutions to the equation, and hence, no real factors of $x^4 + 46$.

If there are no real factors, then a fortiori, there can't be rational factors, either.

8. Feb 20, 2016

### RJLiberator

Ok, beautiful. I understand part 1 very well now. We can show that there is no factors by x^4+46 = 0.
Then, clearly the complex factorization works.

For part two, they ask to completely factor the given polynomial in Z_5.
I believe I have done so by:
p(x) = (4x^2-2x+3)(x+1)

However, in doing so, I haven't used the mod 5 fact at all. All I've done is factor it without any sort of recognition to Z_5 [x].

9. Feb 20, 2016

### Samy_A

You could "recognize" that you work in $\mathbb Z_5$ by checking which other elements of $\mathbb Z_5$ is are roots of the polynomial.

10. Feb 20, 2016

### RJLiberator

You mean multiples of 5?

I could just do:
(9x^2-2x+3)(x+1)

And that would work mod 5.

Is that what you mean by recognizing this element of the question?

11. Feb 20, 2016

### Samy_A

No.
You have a polynomial in $\mathbb Z_5[x]$.
You want to find the zeros in $\mathbb Z_5$ of that polynomial.
There may be fancy methods to do that, but as $\mathbb Z_5$ contains 5 elements, you could just evaluate the polynomial in each of these 5 elements to find all the zeros.

12. Feb 20, 2016

### RJLiberator

Ah, I see what you are saying, so
let x be 1, 2, 3, 4, 5 and see which one makes the equation true:
4x^3+2x^2+x+3 = 0

So we have x = 1, 2, and 4 that make this equation true.

But, er, not sure where to go from here.

We found the roots.

13. Feb 20, 2016

### Samy_A

Now do what one always does: if $a$ is a root of a polynomial $p$ of degree 3, then $p(x)=(x-a)q(x)$, where $q$ is a polynomial of degree 2.
And so on.
Just remember that you are working in $\mathbb Z_5$.
Since you found 3 roots (I found the same 3 roots, so that looks good), you are almost done.

14. Feb 20, 2016

### Staff: Mentor

Not quite -- let x be 0, 1, 2, 3, and 4.
$5 \equiv 0 (\mod 5)$.

15. Feb 20, 2016

### RJLiberator

Excellent correction, Mark.
Indeed. 0 does not work.

p(x) = (x-2)*(4x^2+10x-4)
this should work as well when considering mod 5.

However, I feel I can continually do this over and over.
Is this what you meant by recognizing it?

16. Feb 20, 2016

### Samy_A

Now factor the two other roots too. (I didn't check your second degree polynomial, as we don't need it.)

17. Feb 20, 2016

### RJLiberator

Wopps/ Edit

18. Feb 20, 2016

### RJLiberator

@Samy_A
I finally understand what you meant, it is obvious and important. Sorry I was so dazed this morning.

Here is what I did.
1. Found the roots of the initial polynomial in mod 5. These were 1, 2, and 4.
2. I divided the initial polynomial by one of the roots, I choose (x-1). I was sure to use the mod 5 distinction here.
3. I found that (x-1)(4x^2+x+2) was a factor.
4. I repeated this process for the second term.

I ended up with the answer 4(x+1)(x-1)(x-2) as the completely factored form.