Factoring Polynomials [Abstract Algebra]

[RJLiberator]

Homework Statement

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

Homework 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.

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 can't 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.

My final answers:
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.

 

[Ray Vickson]

Homework Statement

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

Homework 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.

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 can't 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.

My final answers:
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.

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.

 

[Mark44]

Homework Statement

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

It's not clear to me what your notation above means.

Homework 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.

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))

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 $\pi/4, 3\pi/4, 5\pi/4, 7\pi/4$.

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 can't 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.

Right. $x^4 + 46$ can't be factored over the reals.

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.

My final answers:
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.

, post: 5385224, member: 504241"]

Homework Statement

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

Homework 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.

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 can't 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.

My final answers:
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]

 

[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

 

[RJLiberator]

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

 

[Samy_A]

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

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

 

[Mark44]

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

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.

 

[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].

 

[Samy_A]

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].

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.

 

[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?

 

[Samy_A]

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?

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.

 

[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.

 

[Samy_A]

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.

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.

 

[Mark44]

let x be 1, 2, 3, 4, 5

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

 

[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?

 

[Samy_A]

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?

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

 

[RJLiberator]

Wopps/ Edit

 

[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.

I feel great about this answer.

Thank you.