zeros of polynomial. I need to find zeros(including multiple) of $\ x^3+x^2+1$. When I draw graph I notice 1 only elevate the curve. 
So I factored $\ x^3 + x^2$ and had $$ x^2(x+1)$$$= (x+0)^2(x+1)$
So according to what I have read this function have $2$ zeros at $0$ and one at $-1$
Is that right kind of math ?
Thank you. 
EDITED: Sorry. 1,x2 and x3 are bases. I need to know the behavior of them. We have freedom to multiply 1(or any base) with any number. I can multiply 1 with .00000000000001 to bring it almost close to x-axis. –
 A: No that's wrong. If you plug in $x=0$ and $x=-1$ into your original polynomial $x^3+x^2+1$, you get $1$ as result both times, so they are not zeros.
You write that "$1$ only elevate the curve". That's right, but of course elevation by $1$ changes the result of the polynomial (by $1$) and since the zeros of the polynomial are the values of $x$ for which the polynomial becomes $0$, this operation does not preserve zeros.
If you graph the function, you should see that it has local maximum between $x=-1$ and $x=0$ and a local minimum that looks suspicously as though it is around $x=0$. 
Can you calculate the exact position of this local maximum and minimum, using the derivative? 
When you calculate the result if the polynomial at the found extreme positions, you can formulate a region where the zeros of that polynomial can potentially lie, and how many there are (in the real numbers, which I assume for the whole problem)?
The zeros of that polynomial are not some nice, round numbers, you probably need a numerical approximation algorithm to get them to whatever degree you need.
A: Your factoring above used only the expression with $x$ but left alone the value of $1$. When $x^3+x^2+1=0$ we must consider all the terms not only $x^2(x+1)$.
We also can't go far with writing the equation as $x^2(x+1)=-1.$
looking for roots of $f(x)=x^3+x^2+1$ can be accomplished in many ways.
Method $1$ - Plotting
If you plot the curve in a "good interval" you can determine an approximate value for the  root of the function.
If you know one of the roots, you can get the other $2$ roots by polynomial division that would result an equation of degree 2 that could be solved using the Quadratic Formula.
The problem with this approach is that you have to find the interval where the root is and will only get an approximation of the root. The root value shown below is not very accurate. A more accurate value is $x=-1.46557123187676$. Using this value, to get other roots, you need to perform this division:
${(x^3+x^2+1)}/{(x+1.46557123187676)}=x^2 -0.4657x+0.682325$
The right hand side can be solved by the Quadratic Formula, since it is of degree 2. To get the other 2 complex roots. Note that the plot showed only 1 real root. In the values below $i$ stands for $\sqrt{-1}$.

Method $2$ - Using a Numerical Methods
Instead of drawing the curve, you can find a real root by applying Newton's Method. Such methods provide an approximation of the root that gets refined by more iterations into the formula. You need some Calculus to use such methods and you need a starting value to start the process. This works for many types of equations.
Method 3 - Using Cubic Formula.
You can use the Cubic Formula To find all roots. It involves significant arithmetic work to apply, but maybe easier that method 1 above. To use the formula, you don't need to know anything about the roots in advance. The formula will provide values for the 3 root values. using this formula you get the following roots:
$r1=  -1.4655712318767669$
$r2=  0.23278561593838348 +  0.7925519925154489 i$
$r3=  0.23278561593838348 -  0.7925519925154489 i$
This method has more accurate result than plotting.

