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Lesson

You will be shown factoring methods for a variety of polynomials. Practice is essential since, only after you learn to recognize each type, will you be able to factor with ease.

In all you will learn five factoring techniques, and these factoring shortcuts can be used instead of the Factor Theorem and the Rational Root Theorem. Using the new factoring techniques wherever applicable will be of tremendous advantage to you as you continue your study of calculus in university.

Be sure to make notes on how to recognize each type along with the factoring method as you progress through the five cases.

Case 1: Common Factors

You are quite familiar with this factoring technique already as you have been using it since Grade 9.

Write down three cubic polynomials that have no constant term. It doesn't matter which ones you have chosen; you should notice something common about all three. Write down what it is that you notice. Find the roots of your three equations. Notice anything?

Once you have recorded your findings, you may view the correct response.

Case 2: The Sum and Difference of Cubes

Before discussing this type of factoring, let's look first at the difference of two squares.

You should readily recall how to factor something like x2 - 121. Since 121 is (11)2, the factored result is (x - 11)(x + 11). The foil method is a quick way to verify the result.

Now, what about the sum of two squares? Take, for example, x2 + 121.

  • Is this factorable?
  • Can it be factored into (x + 11 )(x +11)?

Checking, by way of the foil method, will soon indicate that this is incorrect. What you must remember is that the square of a binomial always results in a trinomial; there is always a middle term. This is the reason that the sum of two squares will not factor.

You are ready now to examine the sum of two cubes.

In general, p3 +q3 = (p + q)(p2 - pq + q2).

If you think about this for a moment, you will soon realize that it makes sense.

  • Obviously, p + q must be a factor of p3 +q3.
  • As well, the first term of the remaining factor must be p2 to produce the desired p3.
  • The last term must be q2 to obtain q3.
  • As for the middle term, - pq, it is responsible for the cancellation of all other terms, thus giving the desired result p3 +q3.
  • If you are unsure of an answer, you can multiply through to see if it gives the original expression.

Try to factor a few on your own. Factor the following. Record your answers.

  1. x3 + 8
  2. x3 + 64
  3. 27x3 + 216

If you are experiencing difficulty, you might try rewriting the constants as a cube. This will help you identify the values of p and q. Check your results.

The procedure for factoring the difference of two cubes is quite similar.

In general, p3 - q3 = (p - q)(p2 + pq + q2).

Notice that only some of the signs have changed. It is useful to note that the last term of the second factor must be positive to ensure that, when multiplied, -q3 is produced.

Factor the following, recording your answers in your notebook. Again, if you have difficulty, try to rewrite the terms so as to identify p and q.

  1. x3 - 343
  2. 27x3 - 1
  3. x3 - y3

If you are experiencing difficulty, you might try rewriting the constants as a cube. This will help you identify the values of p and q. Check your results.

Read Case 1 and Case 2 as outlined on pages 78 & 79 of your text. Work through the accompanying examples, recording them as you go along. You will note that in all of the presented examples the second factor of the expression does not factor again. However, this is not always the case. It is necessary to examine this expression closely to see if you can indeed factor further.

Case 3: Quartic Expressions Factored as Trinomials

This factoring method works with quartic functions that are written in quadratic form. These are easy to recognize and easy to factor. Below are some tips for identifying quartics that have a quadratic form.

  • The expression consists of three terms.
  • The variable in the first term is the square of the variable in the middle term.
  • The last term is a constant.

An equation of the form x4 - 5x2 + 4 does indeed have a quadratic form. It can be rewritten as:

This, in turn, factors as (x2 - 4)(x2 -1). Each of these factors can be further factored as the difference of two squares as shown below.

x4 - 5x2 + 4 = = (x2 - 4)(x2 -1) = (x - 2)(x + 2)(x - 1)(x + 1)

Some of you may be thinking, "Why not just use the quadratic formula here to find the roots? Then, you will know the factors."

Well, one thing to remember is that the quadratic formula is used to solve quadratic equations, and x4 - 5x2 + 4 =0 is a quartic equation. Since it does have a quadratic form however, the quadratic formula can be used. But, you must realize that in doing so, you are solving for x2, not x. You must, therefore, take the square root of the answers you get to obtain the values for x. Remember, a quartic equation has four roots.

Work through the second example of Case 3 on page 79 of your text. After this task is completed, factor the following. Record your answers.

  1. x4 - 7x2 + 10
  2. 2x4 + 5x2 - 3
  3. 2x4 + 12x2 + 18

Solutions

Case 4: Grouping to find a Common factor

Without even realizing it, you have been using this method for quite some time while factoring quadratics of the form ax2 + bx + c, a 0. using decomposition.

Factor 3x2 - 4x - 4. Record all steps in your notebook. Check your solution.

This process is easily extended to higher degree polynomials. Below are a few tips to help you during this factoring process.

  • An even number of terms must exist to use this method. Do you know why?
  • If four terms are present, consider them as groups of two and look for a common factor in each group.
  • If six terms are present, two different scenarios may be possible. The terms may be considered as three groups of two and your task is to find, and remove, a common factor from each of the three groups. If this doesn't seem to work, consider the six terms as two groups of three. Remove a common factor from each group.
  • Once a common factor is removed from each existing group, the remaining brackets must all contain identical expressions. Remove this expression as a common factor.

Read Case 4 and do the examples given. Record the solutions in your notebook. Note the difference between the second and third examples. After completing this task, factor the expressions below.

  1. x3 - 2x2 + 3x - 6

  2. x5 + x4 + 2x3 +2x2 - 5x - 5

  3. 5x5 -10x4 + 5x3 - 3x2 + 6x - 3

You may verify your solutions.

Case 5: Grouping to get the Difference of Squares

This type of factoring involves decomposition as well as grouping. Before exploring this process, consider the pattern found in perfect square trinomials where a = 1. A few are listed below. Record the pattern. If you are having difficulty finding a pattern, look for the relationship between the linear coefficient and the constant term.

  • x2 - 6x + 9
  • x2 + 10x + 25
  • x2 + 14x + 49
  • x2 - 16x + 64

Once you have recorded the relationship, check your response.

In any equation that will factor in this manner, the middle term is broken into two parts, a perfect square and the correct middle term to create a perfect square trinomial. If you are unable to create these two parts, then the expression cannot be factored using this method.

Read Case 5 and work through the two examples that follow on page 80 of your text. If you are experiencing difficulty, read the following tips on how to recognize and use this factoring technique.

  • See if the polynomial can be grouped in such as way as to give two squares. Try to get it into the form (x + m)2 - n2.
  • To do this, you may have to rewrite the middle term as the sum or difference of two other terms. Keep in mind, your goal is to create "a perfect square trinomial minus a square."
  • Once this is accomplished, factor the difference of two squares.

After studying Case 5, try the examples below for practice.

  1. x4 + 9x2 + 25

  2. x4 - 12x2 + 4

Once you have completed the task, you may check your solutions.

It is imperative that you practice these factoring techniques. To be efficient, you must learn to recognize each type. Factoring with speed and accuracy is essential.

Complete the Focus Questions on page 80 of your text.

Activity

Focus Questions page 80 #'s 40 -42

When you have completed these questions, ask your on-site teacher to get the solutions for you from the Teacher's Resource Binder and check them against your answers. After you do this, if there is something you had trouble with and still do not understand, contact your on-line teacher for help.

Test Yourself

Factor the following:

  1. x3 - 125y3

  2. x4 - x2 + 16
  3. x4 - x2 -72
  4. x3 + 2x2 - 9x -18
  5. 2x3 + 11x2+ 15x

Solutions

1. The difference of two cubes. (x)3 - (5y)3 = (x - 5y)[x2 + 5xy + (5y)2]
= (x - 5y)(x2 + 5xy + 25y2)

2. Case 5 type. It can be grouped to get the difference of 2 squares.

x4 - x2 + 16 = x4 + 8x2 - 9x2 + 16
= (x4 + 8x2 +16) - 9x2
= (x2 + 4)2 - 9x2
 = (x2 + 4 - 3x)(x2 + 4 + 3x)

3. Quartic expression that can be factored as a trinomial. (x2 - 9)(x2 + 8)
= (x + 3)(x - 3)(x2 + 8)

4. Grouping to find a common factor. x3 + 2x2 - 9x -18
= x2(x + 2) - 9(x + 2)
= (x + 2)( x2 - 9)
= (x + 2)(x + 3)(x - 3)

5. Remove a common factor. Then Factor the remaining quadratic expression.

2x3 + 11x2+ 15x = x(2x2 + 11x + 15)
= x(x + 3)(2x + 5)