Factoring polynomials-Can there be fractions?

1. Jun 30, 2013

alingy1

1. The problem statement, all variables and given/known data

I want to factor polynomials. However, I want to know how it is possible to factor polynomials where the linear factors that result do not have integers but rather fractions.

Should we continue factoring if there are fractions, or do we have to stop?

Say we have this factorised form:

(x-0,155266)(x-0,3256)

Is it a factorised form? Or should we find add a parameter to make it integers, i.e.:
a(x-integer1)(x-integrer2)

I have been doing problems involving factoring and I don't get the concept.

2. Jun 30, 2013

darkxponent

The pricipal used for factorising is

x2 - Sx + P.

where S = α+β
and P = α*β

Which only works handy for integers.
For the problem you posted is not easy to factorise. If you start factorising them using the above two equations.

3. Jun 30, 2013

alingy1

So, what do you suggest, do i add a fraction at the beginning of the terms to make the linear factors integers?

4. Jun 30, 2013

SteamKing

Staff Emeritus
If you have (x^2 - 4), you get two nice factors: (x - 2)*(x + 2) = (x^2 - 4)

If you have (x^2 - (9/4)), you get two slightly more complicated factors: (x - (3/2))*(x + (3/2))

What happens if you have (x^2 - 2)?

5. Jun 30, 2013

raunaqrox

I think that if that number, integer or not makes the polynomial 0 then it is a factor .

6. Jun 30, 2013

Ray Vickson

Of course your polynomial above is in factored form! The factored form of a polynomial $p(x)$ is
$$p(x) = a(x-r_1)(x-r_2) \cdots (x-r_n)$$
Here the numbers $r_1, r_2, \ldots, r_n$ are the roots (or zeros) of $p(x)$. There is no requirement that they be integers or fractions, or even real numbers. (Note: some books may say differently, so always check your sources and use the definitions that go along with the course you are taking!)

The more difficult problem is: starting with the polynomial
$$p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0,$$
find the factored form. That requires finding all the roots of $p(x)$.

If the coefficients $a_0, a_1, \ldots, a_n$ are all integers, the question is: does $p(x)$ have rational roots, or not? The rational root theorem handles this question; it gives a finite procedure for deciding the question. If the rational root test passes it discovers/delivers a rational root (so at least one of the roots $r_1, \ldots, r_n$ is rational); if the test fails, none of the roots are rational, and finding even one of the roots may be difficult, possibly requiring numerical solution methods.

7. Jul 1, 2013

HallsofIvy

Staff Emeritus
The question is "What is your purpose in factoring?". Often we factor a quadratic expression to solve the equation= 0. If so, the original form is ideal. As for including a "parameter" as a(x- integer1)(x- integer2) you cannot do that- that would no longer be the same expression. For example, if a quadratic factors as (x- 1/2)(x- 1/3) you seem to be thinking that you can take out the denoinators: (1/6)(x- 1)(x- 1) but that is NOT correct. What you would have would be (1/6)(2x- 1)(3x- 1).