Looks good.jimmycricket said:Homework Statement
Is [tex]f(x)=x^2+1[/tex] irreducible in [tex]\mathbb{F}_2[x][/tex]
If not then factorise the polynomial.
The Attempt at a Solution
[tex]\mathbb{F}_2[x]=\{0,1\}[/tex]
[tex]f(0)=1[/tex]
[tex]f(1)=1+1=0[/tex]
Hence the polynomial is not irreducible
An irreducible polynomial is a polynomial that cannot be factored into polynomials of lower degree with coefficients in the same field. In other words, it cannot be broken down into smaller factors.
One way to determine if a polynomial is irreducible is by using the rational root theorem, which states that if a polynomial with integer coefficients has a rational root, then that root must be a divisor of the constant term. If a polynomial has no rational roots, it is likely to be irreducible. Additionally, checking for repeated roots and using other techniques such as Eisenstein's criterion can also help determine if a polynomial is irreducible.
Yes, an irreducible polynomial can have complex roots. The concept of irreducibility only applies to polynomials with coefficients in a specific field, and complex numbers are a valid field for polynomials. Therefore, a polynomial with complex roots can still be considered irreducible if it cannot be factored into polynomials with coefficients in the complex numbers.
No, not all polynomials of degree 2 or higher are irreducible. For example, x² + 4 can be factored as (x + 2)(x - 2), so it is not irreducible. However, there are certain types of polynomials, such as quadratic polynomials with no real roots, that are always irreducible.
Yes, there are methods for factoring polynomials without using complex numbers. One example is the rational root theorem, which only uses integers. Other techniques such as factoring by grouping and using quadratic formulas can also be used to factor polynomials without involving complex numbers.