3.1: Factoring Polynomials
 Page ID
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In this section, we'll seek to answer the questions:
 What properties of divisibility in \(\mathbb{Z}\) extend to \(F[x]\text{?}\)
 What is an irreducible polynomial? Are there any tools we can use to determine if a given polynomial is irreducible?
One of the most beautiful consequences of an abstract study of algebra is the fact that both \(\mathbb{Z}\) and \(F[x]\) are Euclidean domains. While they are not “the same”, we can expect them to share many of the same properties. In this section, our first goal will be to extend familiar properties from \(\mathbb{Z}\) to \(F[x]\text{.}\) We will also see that particular features of a polynomial (e.g., its degree, or the existence of roots) allows for additional criteria for its irreducibility to be decided.
Since both \(\mathbb{Z}\) and \(F[x]\) have a division algorithm, it is reasonable to expect that, similar to the integers, we can also investigate the greatest common divisor of polynomials. In fact, our method for finding the greatest common divisor of two integers extends nicely to polynomials.
Given \(f(x),g(x)\in F[x]\text{,}\) state a conjecture that gives a means for finding \(\gcd(f(x),g(x))\text{.}\) Prove your conjecture is correct.
Carefully state and prove a Bézoutlike theorem (recall Theorem 1.2.6) for polynomials in \(F[x]\text{.}\)
One of the most useful things we can do with polynomials is evaluate them by “plugging in” elements from our coefficient set (or some superset that contains it) and performing the resulting arithmetic in an appropriate ring. We can make this completely rigorous using the language of functions: given a commutative ring \(R\) and all polynomials \(p(x) = a_0 + a_1 x + \cdots + a_n x^n \in R[x]\text{,}\) we define the function \(p_f : R\to R\) by \(p_f(r) = a_0 + a_1 r + \cdots + a_n r^n \text{.}\) However, we will not belabor this point; instead, we will generally write \(p(r)\) in place of \(p_f(r)\) and appeal to our common notions of evaluating polynomials.
Given a polynomial \(p(x)\in R[x]\text{,}\) we have frequently been interested in finding all \(r\in R\) for which \(p(r) = 0\text{.}\)
Let \(R\) be commutative with identity and suppose \(p(x) \in R[x]\text{.}\) We say \(r\in R\) is a zero or root of \(p(x)\) if \(p(r) = 0\text{.}\)
When considering polynomials with integer coefficients, any rational roots are particularly wellbehaved.
Let \(p(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \in \mathbb{Z}[x]\) with \(a_0,a_n\ne 0\text{.}\) If \(r,s\in\mathbb{Z}\) such that \(s\ne 0, \gcd(r,s)=1\text{,}\) and \(p(r/s)=0\text{,}\) then \(ra_0\) and \(sa_n\text{.}\)
Use Theorem 3.1.1 to find the possible rational roots of \(p(x) = 3  x + x^2 + 5x^3  10x^4  6x^5\text{.}\) Which of the possibilities you found are actually roots? Justify.
Theorem 3.1.1 gave a condition to check to see if polynomials in \(\mathbb{Z}[x]\) had roots in \(\mathbb{Q}\text{.}\) However, the lack of a rational root for a polynomial \(q(x)\in \mathbb{Z}[x]\) is not sufficient to say that a polynomial is irreducible in \(\mathbb{Z}[x]\) according to Definition: Irreducible.
Find a polynomial \(q(x) \in \mathbb{Z}[x]\) that has no roots in \(\mathbb{Q}\) but is nonetheless reducible over \(\mathbb{Z}\).
To simplify matters, we will focus henceforth on polynomials with coefficients in a field. The following theorem is a result that you learned in high school algebra (and have likely used countless times since then), but as with the other familiar topics we have explored so far, it is necessary to formalize prior to continuing.
Let \(F\) be a field, and \(p(x)\in F[x]\text{.}\) Then \(\alpha\in F\) is a root of \(p(x)\) if and only if \(x\alpha\) divides \(p(x)\text{.}\)
Note that while \(F[x]\) is a ring, and we already have a definition of an irreducible element of a ring, we will find it useful to have a ready definition of irreducible in the context of polynomials with coefficients in a field. It is to that task that we now turn.
Given a field \(F\text{,}\) define an irreducible element of \(F[x]\text{,}\) keeping in view Theorem 2.2.8 and Definition: Irreducible.
 Hint

What are the units in \(F[x]\text{?}\)
A polynomial \(f(x)\in F[x]\) is reducible if it is not irreducible.
State a positive definition for a reducible polynomial with coefficients in a field \(F\text{.}\) That is, state a definition which does not refer to the notion of irreducibility.
Every polynomial of degree 1 in \(F[x]\) is irreducible.
A nonconstant polynomial \(f(x)\in F[x]\) of degree 2 or 3 is irreducible over \(F\) if and only if it has no zeros in \(F\text{.}\)
The preceding theorems allow us to explore the (ir)reducibility of polynomials of small degree with coefficients in any field.
Determine which of the following polynomials are irreducible over the given fields. Justify your answer.
 Over \(\mathbb{Z}_2\text{:}\)
 \(x^2 + 1\text{,}\)
 \(x^2 + x\text{,}\)
 \(x^2 +x +1\text{,}\)
 \(x^3 + x^2 + 1\text{,}\)
 \(x^4 + x^2 + 1\text{.}\)
 Over \(\mathbb{Z}_3\text{:}\)
 \(x^2 + 1\text{,}\)
 \(x^2 + x\text{,}\)
 \(x^2 +x +1\text{,}\)
 \(x^2 +x +2\text{,}\)
 \(x^3 + x +1\text{,}\)
 \(x^3 + x^2 + 1\text{,}\)
 \(x^3 + x^2 + x + 1\text{.}\)
As the following theorem illustrates, in \(F[x]\text{,}\) all irreducibles are primes.
Let \(F\) be a field and \(p(x),f(x),g(x)\in F[x]\) such that \(p(x)\) is irreducible and \(p(x)\) divides \(f(x) g(x)\text{.}\) Then \(p(x)\) divides \(f(x)\) or \(p(x)\) divides \(g(x)\text{.}\)
We next state the Fundamental Theorem of Algebra. Despite its name, its proof relies on analytic properties of the real numbers; there is no purely algebraic proof. Moreover, it is not essential for the work we do in following sections, but given its close relationship to the question of factorization, we include it here for completeness.
Every nonconstant polynomial with coefficients in \(\mathbb{C}\) has a root in \(\mathbb{C}\text{.}\)
We conclude with one consequence of the Fundamental Theorem of Algebra.
Every nonconstant polynomial in \(\mathbb{C}[x]\) can be written as a product of linear polynomials.
 Hint

What are the irreducibles in \(\mathbb{C}[x]\text{?}\)
Thus, the multiplicative structure of \(\mathbb{C}[x]\) is straightforward: everything can be factored as a product of linear polynomials. Fields of coefficients like \(\mathbb{C}\) for which this is true are said to be algebraically closed; not all fields satisfy this property. For instance, \(x^2 + 1\in \mathbb{R}[x]\) does not factor into a product of linear polynomials. Consequently, \(\mathbb{R}\) is not algebraically closed.
However, regardless of whether our field is algebraically closed, we have not yet determined that any \(p\in F[x]\) can be factored uniquely into a product of irreducibles, or even that such factorizations into irreducibles exist. In Section 3.2, we do just that.