As we have just seen in the section above, we can use Ruffini's rule as a simple way of finding the answer and remainder of the division of a polynomial divided by the binomial (x - a). In earlier sections we have also seen how to find the numerical value of an algebraic expression in general and also of a specific polynomial.

Example 16.- Find the numerical value of the dividends in exercise 10 when x= -3 and x= 1 respectively. Compare this numerical value to the remainder of the divisions carried out in this exercise.

What do you notice?

In this window you can see the division of cubic polynomials using Ruffini's rule and the numerical value of the dividends for the corresponding values of "a".

If you want to see examples of quadratic polynomials just make the first coefficient (c1) equal to 0.

This window cannot be used for polynomials of a higher degree than 3.

As you have probably seen, the answer can be expressed as:

The remainder theorem:

"If a polynomial P(x) is divided by a binomial (x - a), then the remainder is equal to the numerical value of the polynomial when x = "a", and can be expressed as P(a)".

An important operation when dividing polynomials is factorization, or more specifically, obtaining factors of the type (x-a).

Example 17.- If we calculate the product (x-2)·(x+3) we get the polynomial x² + x - 6, which could be expressed as a product of factors: x² + x - 6 = (x-2 ) · (x+3)

When we manage to express a polynomial as a product of linear binomials, as in the example, or with at least one binomial of this type, we can refer to this as "factorizing the polynomial"

In order to get factors of the type (x - a) we just need to find values of "a" which give us an exact division using Ruffini's rule, i.e. leaving a remainder of 0 and apply:

"Dividend = divisor · quotient + remainder" or D = d · q + r, which gives D = d · q which can be expressed in polynomial terms with the variable x: D(x) = d(x) · q(x). This gives us the polynomial divided into two factors.

You will have noticed that whenever the numerical value is 0, the division of the polynomial by (x - a) is exact (remainder theorem)

If you tried properly you will have found that the numerical value is 0 when x=1 (a = 1) and x = -2. What is the quotient when a = 1?

A very useful rule is: The whole values of "x = a", when the numerical value of a polynomial is zero, are always divisors of the independent term of the polynomial.

This rule makes it easier to find the values of "a". Thus, in example 18 the only possibilities are 1, -1, 2 and -2.

Go back to the last window and try out the rule.

Solving quadratic equations or equations of a higher degree.

Example 19.- In the section above we saw that the following polynomial can be factorized as follows:

2x³ + x² - 5x + 2 = (x - 1) (x + 2) ( 2x - 1)

We also saw that the numerical value of this polynomial when x = 1 and x = -2 is 0, so if the equation is written:

2x³ + x² - 5x + 2 = 0, we know that two solutions are x = 1 and x = -2.

These values of x are called the "roots of the polynomial", which are the solutions of the equation P(x) = 0.

As well as the equation: 2x³ + x² - 5x + 2 = (x - 1) (x + 2) ( 2x - 1) = 0 we also get the following solution as well as the two earlier solutions: 2x - 1 = 0 ; x = 1/2 = 0.5.

(If you would like more background information and practice at solving quadratic or higher degree equations, we recommend working through the unit on equations for the first year of post-compulsory secondary education).