Algebraic Problem:

If A is a real constant then how many solutions does the equation
                  A*x^3+(2-A)*x^2-x-1
have?       
      

First let us define the left-hand side of the equation as a function f(x).

f[x_] = A * x^3 + (2 - A) * x^2 - x - 1

...and check that it is working correctly:

f[1]

f[-2]

A = -7 ; f[-2]

A = -3 ; f[-2]

Clear[A] ;

If we fix a value for A then we can get MATHEMATICA to sketch the graph
of f(x) for us.  
For example if A=-2 we obtain:

A = -2 ;

Plot[f[x], {x, -2, 2}] ;

Clear[A] ;

From this graph we can observe that the equation has three solutions when A=-2.  
Does this agree with the algebraic solution we found?

Next we will sketch the graphs of f(x)for a range of A values between -5 and 5,  
and then animate the results.

Table[     Plot[f[x], {x, -5, 5}, PlotRange→ {-30, 30}] , {A, -5, 5, .5}] ;

Clear[A] ;

By adding modifiers to the basic plot command we can get an output
that's easier to analyze.
And then we can use this to verify the algebraic solutions that we found.

Here's a different way to examine the solutions of our equation:  
The idea is to graph the equation   
                 A*x^3+(2-A)*x^2-x-1
in the  xA-plane.  Then for specific values of A the number of solutions to the
equation will correspond to the number of intersections of the graph with a horizontal
line in the xA-plane.

Table[Show[p2[[NN]], p1, FrameLabel→ {"x-axis", "A-axis"}], {NN, 1, Length[p2]}] ;

Notice that if the original cubic equation were to change we could still run the same MATHEMATICA routines to analyze the solutions, even though a complete algebraic solution may now be very difficult or impossible.

For example, for the quartic equation
            A x^4 + (2-A) x^2 - x - 1 = 0
we would obtain:

Clear[A] ;

f[x] = A * x^4 + (2 - A) * x^2 - x - 1 ;

Clear[f, A] ;

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