From a rational function in the form $ f(x)= \frac{p(x)}{q(x)} $ the following is known.
The numerator polynomial has double zeros $x_{1,2} = 2 $ and a single zero $x_{3} = -2 $ The denominator polynomial has a simple zero at x = -1 and x = 3. Furthermore $ f(0) = - 8$
Im susposed to find out what the function looks like in the as above stated form. My attempt;
$$ f(x) = \frac{(x-2)^2(x+2)}{(x+1)(x-3)} $$ Simplify both nummerator and denominator
$$ f(x) = \frac{x^3-2x^2-4x+8}{x^2-2x-3} $$ This is my final polynomial; the zeros are correct; finding the zeros of the numerator and denominator I get the ones stated in the question. But when I try to check for f(0) I do not get -8. This is what I get
$$f(0) = \frac{8}{3} $$ If I put 0 for all the x's in the numerator and denominator.
Does this mean my function is wrong,or is the way im checking for f(0) is wrong? I am not sure what I am missing here.
Thanks in advance.
I'm going to give an example: If $3$ is a root of say $p(x)=x-3$, then $3$ is also a root of $q(x)=cp(x)=c(x-3)$ for every constant $c$. What you're missing is that, it is not necessary for all polynomials to be monic to have particular roots. So you can multiply by some unknown constant either the numerator or the denominator, and find it afterwards.