Cardan's method for solving cubic equations depends on reducing the general cubic to an equation of the form x3 + px +q = 0. In this problem we explore the changes in the roots of cubic equations as we change the graph so as to learn more about complex numbers. The problem is not about finding actual solutions but about understanding the general case. For completeness we give Cardan's method below.

Consider the general cubic equation
at3 + bt2 + ct + d = 0.
Dividing by a this equation becomes
t3 + b
a
 t2 + c
a
 t + d
a
 = 0.
Making the substitution
t = x - b
3a

gives the equation:
(x - b
3a
 )3 + b
a
 (x - b
3a
 )2 + c
a
 (x - b
3a
 ) + d
a
 = 0.
Expanding this equation is:
x3 - 3( b
3a
 )x2 + 3( b
3a
 )2x - ( b
3a
 )3 + b
a
 (x2 - 2( b
3a
 )x + ( b
3a
 )2) + c
a
 (x - b
3a
 ) + d
a
 = 0.
Clearly the coefficient of x2 is zero and the equation reduces to x3 + px +q=0 where p and q depend on a, b, c and d.

Here, in modern notation, is Cardan's solution of x3 + px = -q .
Notice that (r - s)3 + 3rs(r - s) = r3 - s3 so if r and s satisfy 3rs = p and r3 - s3 = -q then r-s is a solution of x3 + px = -q.
But now s = p/3r so r3 - p3/27r3 = -q, i.e. r6 + qr3 - p3/27 = 0.
This is a quadratic equation in r3, so solve for r3 using the usual formula for a quadratic. Now r is found by taking cube roots and s can be found in a similar way (or using s = p/3r). Then x = r- s is the solution to the cubic.

By the Fundamental Theorem of Algebra every polynomial equation F(z) = 0, where F(z) is a polynomial of degree n has n solutions, some of which may be repeated. So cubic equations have 3 solutions, one of which must be real (why?) and the other two may be real or may be complex.

In school we only consider equations of the form F(z)=0 where the coefficients are real numbers but the methods are the same and it is really no more difficult to consider the general function with complex coefficients.

As we have learnt more about number, so we have been able to find solutions to more equations. In primary school we could find numbers to put in the boxes for equations like [¯]+ 2 = 7, 4×[¯] = 8 and [¯] ×[¯] = 9 and we were really solving the equations x + 2 = 7, 4x = 8 and x2 = 9 without using algebraic notation. We could not solve equations like x + 7 = 2 until we learned about negative numbers. We could not solve equations like 4x = 3 until we learned about fractions (rational numbers) and we could solve equations like x2 = 3 until we learned about irrational numbers. Once we understand complex numbers, which are algebraically very simple, then we are can solve all quadratic equations and we can appreciate that higher order polynomial equations will have real and complex roots.