Find the prime factors of a polynomial?

Question

By using the factor theorem, find the prime factors of the polynomial: x^3-x^2-17x-15

Answer 1

The factor theorem simply states that if you divide a polynomial by one of its factor, you will get a remainder of zero. If a is a root, then (x-a) is a factor:

(a root is a value of x that gives a total of 0 for the polynomial)

p(x) = x^3 - x^2 -17x -15
What the theorem states:
p(x) = (x-a) q(x) exactly (with no remainder)
where q(x) is a second degree polynomial

x^3 - x^2 -17x -15 = (x-a) q(x)

Because your polynomial is third degree, it has at most three real roots (if we were working in complex numbers, we would say it has exactly three roots).

Let us pretend there are three and let us call them a, b and c. It is possible that some of them are equal.

Then, by the theorem, q(x) = (x-b)(x-c) and

x^3 - x^2 -17x -15 = (x-a)(x-b)(x-c)

(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc

Therefore, we know that
(a+b+c) = 1
(ab+ac+bc) = -17
abc = 15

Three unknown, three equations. Should be possible to find an answer.

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Or you could get lucky and find one root (I found x=5 by trial and error):

x^3 - x^2 -17x -15 = (x-5)(x^2 +4x +3)

q(x) being a quadratic should be easier to solve.

The theorem does not give you the answer. It simply shows you that if a root exists, then you can go from p(x), a third degree polynomial, to q(x), a second degree polynomial (much easier to solve).

The theorem works for any degree. If r(x) is a fourth degree polynomial, then if r(x) has a root d, you can find a p(x) such that
r(x) = (x-d) p(x) = (x-d)(x-a) q(x)
and, eventually
r(x) = (x-d)(x-a)(x-b)(x-c)
if the roots exist, of course.