Let's imagine that this type of equation has been simplified as much as possible so that the LHS is a polynomial and the RHS is equal to 0.

Example: x³ + x² - x - 1 = 0

In order to solve this kind of equation we use what is known as "Ruffini's" rule.

Let's suppose that we are principally looking for whole number solutions which we know are the divisors of the independent term of the equation.

Can you find a solution for the example equation above?

Look at the following window. It shows the graph of the function f(x) = x³ + x² - x - 1.

1.- Remember that the solutions of the equation are the points where the graph cuts the X-axis. Can you see a solution?

2.- Change the value of x in the lower part of the window until you find a solution. You could also write in the new values of x manually.

3.- Check your results by solving the equation numerically in your exercise book.

In the following window we can see the same equation, the graph and Ruffini's method being applied for the solution x = 1 as seen earlier.

1.- Note that because the remainder is 0 the value x = 1 must be a solution of the equation.

2.- Remember that the polynomial which corresponds to the LHS of the equation can be broken down into two factors, giving the equation (x-1) (x² +2x + 1) = 0

3.- Solve the equation x² +2x + 1 = 0 (review quadratic equations if necessary). You should have found the other solutions (only one more in this case: x = -1, as you can see in the graph).

Once it has been simplified and arranged correctly, any cubic equation can be expressed in the form:  ax³ + bx² + cx + d = 0 

In general there are between one and three solutions, depending on the factors that we can break the LHS polynomial down into.

The window below initially shows the equation with coefficients:

a = 1, b = -1, c = -1 and d = 1, and has two solutions.

Try changing the values of the parameters a, b, c and d (the values you can use are always whole numbers) to see other types of solutions.

The table also shows Ruffini's rule which allows you to check your solutions numerically.

a) x³ - 2x² + x - 2 = 0

b) x³ + 3x² - x - 3= 0

2.- Check your solutions numerically in your exercise book.

3.- Factorise the LHS of both equations using Ruffini's rule and the quadratic equation solution.

Equations of the fourth or higher degree are dealt with in the same way as above.

For example the equation: x⁴ - x³ - 4x² + 4x = 0 has four solutions which can be seen in the following window.

Check the solutions both numerically and graphically.

Finally, you can change the expression of the equation written at the bottom of the following window (y = LHS once this has been made equal to 0) to find the graphical solution to an equation of any degree.

Note that the equation which is given here is factorised and is of the fifth degree. The five solutions can be seen.