
Date: Sun, 28 Mar 1999 14:05:16 +1000
Subject: Factoring polynomials
My name is Karen
I have been unable to factorise a polynomial equation and was wondering if you could please help. It is level (1012) maths. The polynomial is x^{3} + x^{2}  24x + 36
I have tried a few factorisation methods such as foctorisation by grouping but it won't work this polynomial. Please help.
Hi Karen,
You were right to try factoring by grouping on this and you are right that it doesn't work. One way to attempt to factor a polynomial of degree larger than 2 is to use the ideas of the factor theorem. The factor theorem tells us that if p(x) is a polynomial and p(a) = 0, then x  a is a factor of p(x). For your polynomial, if p(x) = x^{3} + x^{2}  24x + 36 and p(a) = 0 then x^{3} + x^{2}  24x + 36 = (x  a)(some quadratic polynomial). Since the coefficient of x^{3} is 1, the coefficient of x^{2} in the quadratic must also be 1 so if p(a) = 0 then
x^{3} + x^{2}  24x + 36 = (x  a)(x^{2} + bx + c)
The point here is that if you expand the right side then the constant term is ac which must be 36. Thus if a is an integer and p(a) = 0 then a divides 36.
The procedure then is to find factors of the constant term 36 and substitute these values in for x until
you find one that makes p(x) equal to zero. If you know a zero you can
determine a factor. Then use that factor and use either long division or
synthetic division to determine the quotient. In this example p(2) = 0 and division yields
x^{3} + x^{2}  24x + 36 = (x  2)(x^{2}  3x  18)
To complete the factorization you now need to factor x^{2}  3x  18.
An interesting note is that after you find that p(2) = 0 you could keep trying factors of 36 and also find that p(3) = 0 and thus (x  3) is also a factor of p(x) so
x^{3} + x^{2}  24x + 36 = (x  2)(x  3)(x + d). Since (2)(3)(d) = 36, d = 6 and thus
x^{3} + x^{2}  24x + 36 = (x  2)(x  3)(x + 6).
Now check by expanding the right hand side.
Cheers,
Jack and Jeff

