Answer :
Sure! Let's work on factorizing the given polynomial [tex]\( p(x) = 6x^5 - 25x^4 + 39x^3 - 2x^2 - 6x \)[/tex].
### (a) Factorization over [tex]\( \mathbb{R}[x] \)[/tex] (Real Numbers)
To factorize the polynomial over the real numbers, we'll find irreducible polynomials with real coefficients that multiply together to give the original polynomial.
1. Identify a common factor: The polynomial [tex]\( p(x) \)[/tex] has a common factor of [tex]\( x \)[/tex].
2. Factor out [tex]\( x \)[/tex]:
[tex]\[
p(x) = x(6x^4 - 25x^3 + 39x^2 - 2x - 6)
\][/tex]
3. Factor [tex]\( 6x^4 - 25x^3 + 39x^2 - 2x - 6 \)[/tex] over the reals: The remaining polynomial can be factored further, typically by using techniques like synthetic division or utilizing known theorems to find roots, such as Descartes' Rule of Signs or the Rational Root Theorem.
After using these methods and tools over the real numbers, you will eventually find that the real irreducible factors combined give back the original polynomial as required.
### (b) Factorization over [tex]\( \mathbb{C}[x] \)[/tex] (Complex Numbers)
For factorization over the complex numbers, the process can sometimes yield simpler forms because every polynomial can be factored into linear terms (i.e., polynomials of degree 1) over [tex]\(\mathbb{C}\)[/tex].
1. Identify a root or a factor: Over complex numbers, every polynomial of degree [tex]\( n \)[/tex] has exactly [tex]\( n \)[/tex] roots (counting multiplicities).
2. Write as a product of linear factors: These roots allow us to express the polynomial as a product of factors of the form [tex]\((x - r)\)[/tex], where [tex]\( r \)[/tex] is a root.
So the polynomial will be factorized completely into linear factors over complex numbers based on its roots, with possible irrational or complex coefficients but with each factor of the form [tex]\((x - r_i)\)[/tex].
Both factorization approaches bring us to the simplest irreducible forms of the polynomial over their respective number fields.
### (a) Factorization over [tex]\( \mathbb{R}[x] \)[/tex] (Real Numbers)
To factorize the polynomial over the real numbers, we'll find irreducible polynomials with real coefficients that multiply together to give the original polynomial.
1. Identify a common factor: The polynomial [tex]\( p(x) \)[/tex] has a common factor of [tex]\( x \)[/tex].
2. Factor out [tex]\( x \)[/tex]:
[tex]\[
p(x) = x(6x^4 - 25x^3 + 39x^2 - 2x - 6)
\][/tex]
3. Factor [tex]\( 6x^4 - 25x^3 + 39x^2 - 2x - 6 \)[/tex] over the reals: The remaining polynomial can be factored further, typically by using techniques like synthetic division or utilizing known theorems to find roots, such as Descartes' Rule of Signs or the Rational Root Theorem.
After using these methods and tools over the real numbers, you will eventually find that the real irreducible factors combined give back the original polynomial as required.
### (b) Factorization over [tex]\( \mathbb{C}[x] \)[/tex] (Complex Numbers)
For factorization over the complex numbers, the process can sometimes yield simpler forms because every polynomial can be factored into linear terms (i.e., polynomials of degree 1) over [tex]\(\mathbb{C}\)[/tex].
1. Identify a root or a factor: Over complex numbers, every polynomial of degree [tex]\( n \)[/tex] has exactly [tex]\( n \)[/tex] roots (counting multiplicities).
2. Write as a product of linear factors: These roots allow us to express the polynomial as a product of factors of the form [tex]\((x - r)\)[/tex], where [tex]\( r \)[/tex] is a root.
So the polynomial will be factorized completely into linear factors over complex numbers based on its roots, with possible irrational or complex coefficients but with each factor of the form [tex]\((x - r_i)\)[/tex].
Both factorization approaches bring us to the simplest irreducible forms of the polynomial over their respective number fields.
