# Thread: Advanced Factoring Question

1. ## Advanced Factoring Question

Hi all! I am actually trying to evaluate an integral but first I need to factor this expression.

$30x^3+23x^2-29x+6$

Now since I have a solutions manual, I already know that the factorization of this is:

$(2x+3)(3x-1)(5x-2)$

So the rational zeros are $\frac{-3~}{2},~\frac{1}{3}~,\frac{2}{5}$. 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 $\frac{a}{b}$ for some $a,~b \in \mathbb{Z}.$ Does that ALWAYS mean that $(bx -a)$ is a factor of the expression?

Thanks for any help!
James

2. Originally Posted by james121515
Hi all! I am actually trying to evaluate an integral but first I need to factor this expression.

$30x^3+23x^2-29x+6$

Now since I have a solutions manual, I already know that the factorization of this is:

$[(2x+3)(3x-1)(5x-2)$

So the rational zeros are $\frac{-3~}{2},~\frac{1}{3}~,\frac{2}{5}$. 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 $\frac{a}{b}$ for some $a,~b \in \mathbb{Z}.$ Does that ALWAYS mean that $(bx -a)$ is a factor of the expression?

Thanks for any help!
James
Surely you're making extra work for yourself? Unless this expression is a denominator and you need to use partial fractions integrating it direct would make more sense

The only way I'd be able to solve this is to guess values or use an iterative solution to a 0

3. Check out this : Polynomial long division - Wikipedia, the free encyclopedia

Basically, once you have found one rational root $\frac{a}{b}$ 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 $(bx - a)$, until you found all your factors (or until the remaining polynomial is small enough to be solved by an easy process : quadratic, or cubic).

4. 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

5. 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.

6. Originally Posted by james121515
Yes I am indeed do a partial fraction integration
So why did you just ‘drop us into’ the middle of the question?
Why did you not post the original integration question?
If this is simply a question of ‘partial fractions’, then the original problem is necessary.

7. i was curious as to the concepts behind factoring polynomials, hence why I posted in the algebra forum. if I had a question about the calculus I would have posted in that forum.

8. Originally Posted by james121515
i was curious as to the concepts behind factoring polynomials, hence why I posted in the algebra forum. if I had a question about the calculus I would have posted in that forum.
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.

9. Originally Posted by james121515
Hi all! I am actually trying to evaluate an integral but first I need to factor this expression.

$30x^3+23x^2-29x+6$

Now since I have a solutions manual, I already know that the factorization of this is:

$(2x+3)(3x-1)(5x-2)$

So the rational zeros are $\frac{-3~}{2},~\frac{1}{3}~,\frac{2}{5}$. correct?
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

10. Yes, synthetic division works with any number, even fractions or irrational numbers. But if you have a root like 2/3, you are probably better off doing ordinary polynomial division with 3x-2.