# Irreducible Polynomials

1. Sep 9, 2010

### beetle2

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

Write $P(x) = x^3+2x+3$ as the product of Irreducible Polynomials over $Z_5$

2. Relevant equations

Polynomial division

3. The attempt at a solution

I start out by taking out a factor of $x+3$

That is

$x+3 \div x^3+2x+3$

I get $P(x) = x^2-3x+1$ which has zero remainder mod 5.

Is the product of irreducible polynomial $(x+3) (x^2-3x+1)$

or do I reduce $P(x) = x^2-3x+1$ by taking out a factor of x+1 ie

$x+1 \div x^2-3x+1$

I know the irreducible polynomials coefficients should add up to the original degree ,So I have one with degree 1 and the second with degree 2.

am I on the right track?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 9, 2010

### Office_Shredder

Staff Emeritus
Well, a product of irreducible polynomials requires all your polynomials to be irreducible.

You're on the right track, the question now is: is x+3 irreducible, and is x2-3x+1 irreducible?

3. Sep 9, 2010

### beetle2

I evaluated

$x+1 \div x^2-3x+1$

which is $P(X)= x-4$ zero remainder mod 5

So I have three irreducible Polynomials whose degrees add to three ie

$(x+3)(x+1)(x-4)$

Hows that look

4. Sep 9, 2010

### beetle2

Is there a way to check that my answer is right?

5. Sep 9, 2010

### Dick

Multiply your product out and reduce the coefficients mod 5.

6. Sep 9, 2010

### beetle2

I multiply it out and get

$x^3-13x-12$ which is $x^3-3x-2$mod 5 so I'm doing something wrong.

7. Sep 9, 2010

### Dick

Don't forget 2=(-3) and 3=(-2) mod 5.

8. Sep 9, 2010

### beetle2

I think I need some more practice, It can get confusing just doing ordinary polynomial division without having modulo as well
$x^3+2x+3$