What should you always look for first when attempting to factor a polynomial completely?

Answer 1

This is a good question and does not have a general answer!

Let's call p(x) a polynomial in x of degree n>=0 an expression of the form.

(If n=0 the polynomial is called constant)

p(x) = Sum(a_i * x^i, i=0..n),

this is a sum of terms a_i * x^i where i ranges from 0 to n.

A factor of a polynomial p(x) is another polynomial q(x) that divides it. (Since we can define division among polynomials, it turns into a ring. But that's beyond this answer). For instance if there is a polynomial r(x) such that p(x) = q(x)*r(x), then p(x)/q(x) = r(x). This is pretty straightforward. Note that divisors of a polynomial can have any degree, hence do not have to have degree 1.

Mathematicians have proved that polynomials of degree 1, 2, 3, 4, 6, 8 can be factored with reasonable methods, but if the degree is 5, 7 or higher than 8 there is no general method. I'll give another positive result later on.

The reason for this is the following.

Factoring polynomials has a relation with finding its roots.

A root (or zero) of a polynomial p(x) is a value of x such that p(x) = 0.

Now suppose that p(x)=0 and p(x) can be factored as p(x) = q(x)*r(x), then it follows that q(x) = 0 or r(x) = 0 and we found a zero of one of its divisors. We can turn this around too: if p(a) = 0 for some a, then there are polynomials q(x) and r(x) such that p(x)/q(x) = r(x). In particular q(x) = x-a can be taken. (I won't prove it here.)

This gives you one method to find a factor of a polynomial: find a zero of the polynomial.

The main theorem of algebra is this:

Every polynomial has a complex zero.

A complex number is of the form x + i.y, where

i = (-1)^(1/2) is the root of X^2 + 1.

A complex number is always the root of a second degree polynomial with real coefficients a_i (see definition of polynomial). (I won't prove that here).

This gives you another method to find a factor: find a complex root of p(x) and find the second degree polynomial that it is a root of. This will be a factor of p(x).

Example: Factor X^4 + 1.

Answer: [X^2 - X*{2^(1/2)}+ 1]*[X^2 + X*{2^(1/2)}+ 1]

Note that only real numbers appear in this expression!

To check the answer, just multiply the factors to arrive at the original polynomial.

I like to give you the derivation of this, since it will show you how the complex roots work together to find the factors of X^4 + 1.

Note that the factors of this polynomial lie on a circle of radius one, and form a square with points {(1/2)*2^(1/2)*(1+i), (1/2)*2^(1/2)*(1-i), (1/2)*2^(1/2)*(-1+i); (1/2)*2^(1/2)*(-1-i)}.

For better understanding, copy this and change the expression (1/2)*2^(1/2) as half of the square root of 2.

This actually means that

X^4 + 1 =

[X - (1/2)*2^(1/2)*(1+i)]*

[X - (1/2)*2^(1/2)*(1-i)]*

[X - (1/2)*2^(1/2)*(-1+i)]*

[X - (1/2)*2^(1/2)*(-1-i)]

Now multiply the first two factors and the last two factors.

First note that (1+i)*(1-i) = 1 - i^2 = 1 - (-1) = 2 to find

X^4 + 1 =

[X^2 - X*{(1/2)*2^(1/2)*(1-i)+(1/2)*2^(1/2)*(1+i)}+ 2*{(1/2)*2^(1/2)}^2]

[X^2 - X*{(1/2)*2^(1/2)*(-1+i)+(1/2)*2^(1/2)*(-1-i)}+ 2*{(1/2)*2^(1/2)}^2]

=

[X^2 - X*{2^(1/2)}+ 1]

[X^2 + X*{2^(1/2)}+ 1]

Answer 2

The GCF. Or the greatest common factor.