Posted by Yoona on Saturday, November 5, 2011 at 10:28pm.
For (a), do some trial solutions by substituting x=factors of 20, such as
f(1),f(2),f(5),f(10), etc.
If you find one (there is one distinct root, and two coincident roots). The coincident roots are positive (as can be guessed by the Descartes rule of signs).
Once you found one, you just have to divide f(x) by the first zero to get a quadratic, which can then be factored by standard methods.
If you need further help, please post your attempts.
I'm in AP calc. We still didn't learn what descartes rule is.. Can you be more explisive please?
Descartes rule of signs is probably not important in this context,but here it is:
If the polynomial is arranged in descending order of the powers of the (single) variable, the number of change of signs indicates the maximum number of positive zeroes.
In the given case, there are two changes of sign, so the number of positive roots is either two or zero.
(see following link for a more detailed version)
http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs
In solving for rational roots of a cubic, one of the ways is to try all possible combinations of the following as zeroes:
(factors of the constant term)/(factors of the leading coefficient).
In the case in point, we need to try as zeroes all of:
(±20,±10,±5,±2,±1)/(±2,±1).
At the beginning it looks like it's a lot of work, but as you get more experienced, it will be very straightforward. You just have to get organized.
For example, check the values of (start with small values, they are usually more likely to be zeroes):
f(1/2)
f(-1/2)
f(1)
f(-1)
f(2)
f(-2)
f(5)
f(-5)
f(5/2)
f(-5/2)
f(10)
f(-10)
f(20)
f(-20)
These are all the possible rational roots to try.