# given the equation x^3-2x^2+x-3=0, what are the possible roots?please help me understand this, i looked at the book and it left me so confused. i have been trying to figure it out all day and i...

given the equation x^3-2x^2+x-3=0, what are the possible roots?

please help me understand this, i looked at the book and it left me so confused. i have been trying to figure it out all day and i cant.:( help me, please expalin the steps and how to get the answer. thank you very much and your help is appreciated

embizze | Certified Educator

Given `x^3-2x^2+x-3=0` , we are asked to find the possible roots.

(1) From the rational root theorem we know that the only possible rational roots are `+-1,+-3`

** The rational root theorem says that any rational roots are in the form `p/q` where p divides the constant term, and q divides the leading coefficient. Thus p can be `+-1,+-3` (divisors of -3) and q can be `+-1` (divisors of 1)**

(2) We can use synthetic division to check each -- there is a rule associated with Descartes rule of signs: if the result row of the synthetic division is all positive, then there are no roots larger than the divisor, while if the signs alternate there are no roots smaller than the divisor.

Checking 1 we get:

1  |  1   -2   1   -3
-----------------
1   -1   0   -3

1 is not a root, and there are no roots smaller than 1.

Checking 3 we get:

3  |  1   -2   1   -3
----------------
1   1    4   9

3 is not a root, and there are no roots larger than 3.

Thus there are no rational roots.

(3) Checking the graph we see that the only real zero occurs between x=2 and x=3; using a graphing utility we find the zero to be approximately 2.1745594

(4) By Descartes rule of signs there are either 3 positive roots or 1 positive root, and there are no negative roots.

The graph: