Hi all! I am actually trying to evaluate an integral but first I need to factor this expression.
Now since I have a solutions manual, I already know that the factorization of this is:
So the rational zeros are . correct?
However, my solution manual does not explain the process of factoring this. How exactly do you factor this? My guess is that you can do a rational root test and find the rational zeros by trial and error with synthetic division. But, first of all, is it even possible to perform synthetic with fractions? And if so, and say you find that one rational root of some polynomial is for some Does that ALWAYS mean that is a factor of the expression?
Thanks for any help!
James
Check out this : Polynomial long division - Wikipedia, the free encyclopedia
Basically, once you have found one rational root by trial (or by just plotting the polynomial and looking at the roots), you can perform a long division on your polynomial with the factor , until you found all your factors (or until the remaining polynomial is small enough to be solved by an easy process : quadratic, or cubic).
Yes I am indeed do a partial fraction integration here, but at the same time I really want to understand the concepts behind factoring. This is, in fact, the denominator of my function I want to integrate. But it's not the integration I'm trying to understand here, it's the factoring.
Thanks for your reply e,
James
Hi James,
the Newton-Raphson method will find roots for you,
but will give you decimal rather than rational form.
Otherwise, since your solutions are rational,
you can proceed with a trial and error process as follows.
Your function is a cubic, which you can break into a pair of quadratic and linear factors using (ax^2+bx+c)(dx+e).
This is a first stage of two stages, as finally you will factorise the quadratic but that is much simpler.
Hence, multiplying out (ax^2+bx+c)(dx+e) gives
ad=30, ae+bd=23, be+cd=-29, ce=6.
The major clue is that the co-efficient of x is -29.
By a trial and error process, you discover that if b=-11,
you find a=15, d=2, c=2, e=3.
(15x^2-11x+2)(2x+3)=(3x-1)(5x-2)(2x+3).
Also, if a/b is a root, or solution of f(x)=0, then this means f(a/b)=0,
hence x-(a/b)=0 for x=(a/b).
Since factorising is the process of writing the function as a multiplication of products,
then x-(a/b) must be one of the factors, which means bx-a =0 as b(0)=0.
It can often be the case that cubic functions appearing in an integrand do not require factorising. Since this particular cubic is not one that exemplifies by-hand factorisation, Plato is quite rightly justified in questioning why the whole problem has not been posted. It is the experience of many senior members that when a student posts only part of the problem, the lack of context causes problems and leads to a waste of time.
The rational roots theorem tell you that if this has rational roots they are of the form of a factor of the constant term divided by a factor of the coefficient of the highest order term.
Hence if it has rational roots they are among +/- the following
1, 1/2, 1/3, 1/5, 1/10, 1/15, 1/30, 2, 2/3, 2/5, 2/15, 3, 3/2, 3/5, 3/10, 6, 6/5
Now we just try these to see which if any are roots.
CB