Irreducible Polynomials

Try this

With $s = 0$ , all the coefficients of $f (x) = x^{4} + x^{3} + x^{2} + x + 1$ are $1$ . No prime divides $1$ , so the criterion has nothing to work with.
Drag $s$ to $1$ : the expression becomes $f (x + 1)$ , with coefficients $1, 5, 10, 10, 5$ . Choose $p = 5$ , and all three conditions hold.
A shift does not change whether a polynomial factors: if $f (x) = g (x) h (x)$ , then $f (x + 1) = g (x + 1) h (x + 1)$ , and the other way round too. So if $f (x + 1)$ is irreducible, so is $f (x)$ .
Click “Ask the kernel” to have the computer factor $f (x)$ directly, and check.

Irreducible polynomial: a polynomial that, over the rationals, cannot be written as a product of two polynomials of lower degree. It plays the role that primes play among the integers.

To decide whether a number is prime, you can try dividing by one number after another. For polynomials there is no such direct way to decide irreducibility: “I could not split it” is not the same as “it cannot be split”. Eisenstein's criterion gives a proof:

Take an integer polynomial $a_{n} x^{n} + \dots + a_{1} x + a_{0}$ . If there is a prime $p$ such that ① $p$ does not divide the leading coefficient $a_{n}$ ; ② $p$ divides every other coefficient; ③ $p^{2}$ does not divide the constant term $a_{0}$ , then it is irreducible over the rationals.

Predict first

For $x^{2} + 1$ , no prime $p$ satisfies the conditions of the criterion. Does that show $x^{2} + 1$ is reducible?

Yes, it is reducibleNo: the criterion is only a sufficient condition

Choose an answer first; then you can see the result.

Why

Example: is x⁴ + 6x + 3 irreducible?

Step 1 / 5

Try p = 2 first
$2$ divides $0$ , $0$ and $6$ , but not the constant term $3$ . Condition ② fails, so try another.

$p = 2 : 2 ∤ 3$
Now try p = 3
Why that is enough (1)
Why that is enough (2)
A contradiction

What xⁿ − 1 looks like when factored completely

$x^{n} - 1$ is never irreducible (it always has the factor $x - 1$ ). Factor it completely over the rationals, and each irreducible factor you get is called a cyclotomic polynomial, written $Φ_{d}$ . In the picture below, the roots labeled with the same number $d$ belong to the same factor $Φ_{d}$ .

Drag

n

. Which numbers

d

appear on the roots? How are they related to

n

Predict first

Factor $x^{12} - 1$ completely over the rationals. How many factors are there?

Choose an answer first; then you can see the result.

Remove one condition

Condition ③ of the criterion looks the least important. Remove it and see.

Break it

Counterexample lab: remove one condition

Conclusion: A polynomial satisfying ① and ② is irreducible

$x^{2} + 4 x + 2$ with $p = 2$ : all three conditions hold, so it is irreducible. Click “Ask the kernel” to check.

Practice

Exercise 1Basic

To prove that $x^{3} + 6 x^{2} + 9 x + 3$ is irreducible with Eisenstein's criterion, which prime $p$ should you use?

Exercise 2Basic

Factor $x^{6} - 1$ completely over the rationals. How many factors are there?

Exercise 3Advanced

$x^{4} + x^{2} + 1$ has no rational root. Is it irreducible? If not, factor it.

Exercise 4Challenge

Of the four polynomials $x^{4} + 4$ , $x^{4} + 2$ , $x^{4} + 1$ and $x^{4} - 1$ , how many are irreducible over the rationals?

Try this

Why

Example: is x⁴ + 6x + 3 irreducible?

What xⁿ − 1 looks like when factored completely

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

The criterion does not apply = reducible?

No rational root = irreducible?