Typing math like that uses something like LaTex - see
ftp://ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf
for example for some uses. To see some actual code, just do a reply to one of the posts with the math symbols in it, like this one for example.
\(\displaystyle This\space is\space an\space example of a partially LaTex formatted post\space \sqrt{2437}\)
To get back to the problem: We want to find the zeros of f(x) where
f(x) = \(\displaystyle \frac {5 x^2 + 4 x - 1}{2 \sqrt {x+1}}\)
You started out correctly by finding the zeros of
g(x) = \(\displaystyle 5 x^2 + 4 x - 1\)
There is a theorem which says if p(x) is a polynomial
p(x) = \(\displaystyle a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0\)
with \(\displaystyle a_n \ne 0\) and zeros
{\(\displaystyle x_0, x_1, x_2, ..., x_n\)},
the p(x) can be written as
p(x) = \(\displaystyle a_n (x-x_0) (x-x_1) (x-x_2) ... (x-x_n)\)
Applying that to our case here for g(x), whose zero are -1 and 0.2, we have
g(x) = \(\displaystyle 5 (x + 1) (x - 0.2)\)
so f(x) can be written
f(x) = \(\displaystyle \frac {5 (x + 1) (x- 0.2) }{2 \sqrt {x+1}} = \frac{5}{2} \sqrt {x + 1}\space (x- 0.2) \)
So the zeros of f(x) are -1 and 0.2.
To show the importance of first solving for the zeros of g(x), suppose our function f(x) had been
\(\displaystyle \frac {5 x^2 + 4 x - 1}{2 (x+1)}\).
What would have been the zeros then.
