Factoring over the Reals and the Complex Numbers

Try this

$n = 4$ , over the rationals: $x^{4} - 1$ splits into $3$ factors, and one of them, $x^{2} + 1$ , is quadratic. In the picture it is in charge of the two roots at the top and bottom, $i$ and $- i$ .
Switch to the reals: still $3$ . The two roots of $x^{2} + 1$ are not on the real axis, so real numbers do not help.
Switch to the complex numbers: $x^{2} + 1$ splits too, giving $4$ linear factors, one for each root.
Set $n$ to $8$ . Over the rationals there is a quartic factor $x^{4} + 1$ ; over the reals it splits into two quadratics, with $2$ in their coefficients. Look at the picture: each quadratic is in charge of a pair of roots that are mirror images in the real axis.

In three sentences:

Over the complex numbers: a polynomial of degree $n$ splits into exactly $n$ linear factors.
Over the reals: when you factor completely, you only get linear factors and quadratic factors with no real roots.
Over the rationals: irreducible factors can have any degree (for example $x^{4} + 1$ ).

For a quadratic $a x^{2} + b x + c$ (with rational coefficients), the discriminant $Δ = b^{2} - 4 a c$ tells you everything:

$Δ$	Over the rationals	Over the reals	Over the complex numbers
the square of a rational number	factors	factors	factors
positive, but not the square of a rational	does not	factors	factors
negative	does not	does not	factors

Predict first

Over which numbers, at the least, can $x^{2} - 2$ be factored?

The rationals are enoughIt needs the realsIt needs the complex numbers

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

Why

Why do roots off the real axis always pair up into a quadratic with real coefficients?

Let $ω$ be a root on the unit circle at angle $θ$ to the positive real axis. Its mirror image in the real axis is written $\overset{ω}{ˉ}$ (the conjugate). Multiply the two linear factors:

(x - ω) (x - \overset{ω}{ˉ}) = x^{2} - (ω + \overset{ω}{ˉ}) x + ω \overset{ω}{ˉ} = x^{2} - 2 cos θ x + 1 .

In $ω + \overset{ω}{ˉ}$ the imaginary parts cancel exactly, and $ω \overset{ω}{ˉ}$ is the square of the distance to the origin. Both coefficients are real.

x⁸ − 1: the more numbers, the finer the split

Step 1 / 5

Over the rationals
$x^{8} - 1 = (x^{4} - 1) (x^{4} + 1) = (x - 1) (x + 1) (x^{2} + 1) (x^{4} + 1)$ . Four factors; $x^{4} + 1$ is in charge of the four “diagonal” roots.

$(x - 1) (x + 1) (x^{2} + 1) (x^{4} + 1)$
Switch to the reals
This is really adding a term
As far as the reals go
Over the complex numbers

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: $x^{2} + 1$ cannot be split

Over the reals, the $x^{2} + 1$ in $x^{4} - 1$ stays in one piece: its two roots $\pm i$ have to stay together.

Practice

Exercise 1Basic

Factor $x^{4} - 9$ completely over the reals. How many linear factors are there in the result?

Exercise 2Basic

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

Exercise 3Advanced

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

Exercise 4Challenge

Over the reals, $x^{4} + 1 = (x^{2} + a x + 1) (x^{2} - a x + 1)$ with $a > 0$ . Find $a^{2}$ .

Factoring over the Reals and the Complex Numbers

Try this

Why

x⁸ − 1: the more numbers, the finer the split

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Not reading which numbers the problem allows

Negative discriminant = can never be split?