How do you know when you have completely factored a polynomial?

1 Answer
Jun 28, 2018

Answer:

It depends...

Explanation:

The shortest, but slightly misleading answer might be: when all of the factors are linear.

What is missing from the question and the answer is to say what set of numbers the coefficients of the factors come from. If the coefficients of the given polynomial are any ordinary numbers - i.e. anything from integers to irrational and/or complex numbers - then in theory the polynomial will factor into a product of linear factors with complex zeros corresponding to its zeros.

For example:

#x^3-x^2+2x-2 = (x-1)(x-sqrt(2)i)(x+sqrt(2)i)#

If you wanted to "completely factor over the reals", then you would not be able to reduce this polynomial to a product of linear factors. Instead you would have to stop at:

#x^3-x^2+2x-2 = (x-1)(x^2+2)#

In theory, any polynomial in one variable with real (e.g. integer) coefficients can be factored as the product of linear and/or quadratic factors where the quadratic factors are irreducible over #RR#.

For example, despite having no real zeros, the quartic #x^4+1# is factorable over the reals as:

#x^4+1 = (x^2-sqrt(2)x+1)(x^2+sqrt(2)x+1)#

To tell whether a quadratic polynomial with real coefficients is irreducible over #RR#, you can examine its discriminant. The discriminant #Delta# of a quadratic of the form:

#ax^2+bx+c#

is given by:

#Delta = b^2-4ac#

The quadratic is factorable (over #RR#) if and only if #Delta >= 0#.

In practice a polynomial factorisation is not necessarily expressible in terms of elementary functions (i.e. #n#th roots, etc). For example:

#x^5+4x+2 = (x+a)(x^2+bx+c)(x^2+dx+e)#

for some real numbers #a, b, c, d, e#, but none of these coefficients is expressible in terms of ordinary arithmetic expressions and #n#th roots.