Finding a parameter of a function I have a function: $f(x)=-\frac{4x^{3}+4x^{2}+ax-18}{2x+3}$ which has only one point of intersection with the $x$-axis.
How can i find the value of $a$?
I tried polynomial division and discriminant, but it didn't help me.
 A: I played around with the numerator of this function in Mathematica.  Lots of values of $a$ give one root, but only one value of $a$, $a=-15$, gives two distinct real roots:

The double root is at $x=-3/2$, the single at $x=2$.  I suspect this is what you were looking for.  In this case, 
$$f(x) = (2 x+3) (x-2)$$
A: This may be a little longer than needed, but I wanted to show my thought process in finding the solution.
First of all, let's ignore the minus sign, it doesn't make any difference. Call $P(x) = 4x^3+4x^2+ax-18$ and $Q(x) = 2x+3$. Note that if $x \neq -\frac32$, then $Q(x) \neq 0$ and so we can pretty much ignore it when looking for roots.
Suppose $x=-\frac32$. Then $P(-\frac32) = 0$ if and only if $a = -15$ (you can check this by direct evaluation). We can factor $P$ in this case: $P(x) = 4(x+\frac32)^2(x-2)$. $Q$ can also be written as $Q(x) = 2(x+\frac32)$, so we can cancel $x+\frac32$ and find that while the function is not defined at $x=-\frac32$, its limit is $0$. $f(x)$ is also zero at $x = 2$, so if you count $\frac32$ as an intercept then $a=-15$ isn't a solution, while if you count it then it is.
Now suppose $x\neq -\frac32$. The problem reduces to $P(x) = 4x^3+4x^2+ax-18 = 0$. This is cubic equation, so it's gonna be pretty hard to solve it directly. Instead, let's look at the derivative: $P'(x) = 12x^2+8x+a$. If we set that equal to zero we get $x = \frac{-2\pm\sqrt{4-3a}}{6}$. If $a \gt \frac43$, this has no roots and is always positive. Therefore $P$ is increasing and it has only one root.
If $a = \frac43$ then $P'$ has a double root. $P$ is still increasing (though not strictly so), and therefore has only one root.
It seems like if $-15 \lt a \lt \frac43$ there's only one root, and if $a \lt -15$ there's three. I don't know how to prove this yet, so I'm posting this incomplete answer hoping it will be helpful.
A: I think this might work. The horizontal asymptote of this function is at $x =\frac{-3}{2}$. So whenever this division yields $0$ remainder this asymptote would not work since this means that we have some form of $(2x+3)(\alpha x^{2}+\beta x+ \theta)$. If this is the case than asymptote would not work out. So by polynomial division when $a =-15$ there is no remainder, however as long as $a>-15$ there is remainder and asymptote will work.
