Common factor by grouping of terms examples, exercises

The common factor by grouping of terms is an algebraic procedure that allows you to write some algebraic expressions in the form of factors. To achieve this objective, it is first necessary to group the expression properly and observe that each group thus formed has, in effect, a common factor.

Applying the technique correctly requires some practice, but in a short time you can master it. Let's first look at an illustrative example described step by step. Then the reader can apply what they have learned in each of the exercises that will appear later.

For example suppose you need to factor the following expression:

2x^two + 2xy - 3zx - 3zy

This algebraic expression consists of 4 monomials or terms, separated by + and - signs, namely:

2x^two, 2xy, -3zx, -3zy

Looking closely, x is common to the first three, but not the last, while y is common to the second and fourth, and z is common to the third and fourth..

So in principle there is no common factor to the four terms at the same time, but if they are grouped as will be shown in the next section, it is possible that one will appear that helps to write the expression as the product of two or more factors.

Article index

• 1 Examples
• 2 Important questions about the common factor by grouping
• 3 Exercises
  • 3.1 - Exercise 1
  • 3.2 - Exercise 2
• 4 References

Examples

Factor the expression: 2x^two + 2xy - 3zx - 3zy

Step 1: Group

2x^two + 2xy - 3zx - 3zy = (2x^two + 2xy) + (-3zx - 3zy)

Step 2: Find the common factor of each group

2x^two + 2xy - 3zx - 3zy =

= (2x^two + 2xy) - (3zx + 3zy) =

= 2x (x + y) - 3z (x + y)

Iimportant: the negative sign it is also a common factor that must be taken into account.

Now notice that the parentheses (x + y) is repeated in the two terms obtained by grouping. That is the common factor that was being sought.

Step 3: Factor the entire expression

2x^two + 2xy - 3zx - 3zy = (x + y) (2x - 3z)

With the previous result, the goal of factoring has been reached, which is none other than transforming an algebraic expression based on additions and subtractions of terms, into the product of two or more factors, in our example, of: (x + y) y (2x - 3z).

Important questions about the common factor by grouping

Issue 1: How to know that the result is correct?

Answer: The distributive property is applied to the result obtained and after reducing and simplifying, the expression thus achieved must match the original, if not, there is an error.

In the previous example, we work in reverse with the result, to check that it is correct:

(x + y) (2x - 3z) = 2x^two -3zx + 2xy - 3zy

As the order of the addends does not alter the sum, after applying the distributive property all the original terms are returned, signs included, therefore, the factorization is correct.

Question 2: Could it have been grouped in another way?

Answer: There are algebraic expressions that allow more than one form of grouping and others that do not. In the selected example, the reader can try other possibilities on his own, for example grouping like this:

2x^two + 2xy - 3zx - 3zy = (2x^two- 3zx) + (2xy - 3zy)

And you can check that the result is the same as it was obtained here. Finding the optimal grouping is a matter of practice.

Question 3: Why is it necessary to take a common factor from an algebraic expression?

Answer: Because there are applications in which the factored expression makes calculations easier. For example, suppose you want to do 2x^two + 2xy - 3zx - 3zy equal to 0. What would be the possibilities?

To answer this question, the factored version is much more useful than the original development in terms. It is posed like this:

(x + y) (2x - 3z) = 0

One possibility that the expression is worth 0 is that x = -y, regardless of the value of z. And the other is that x = (3/2) z, regardless of the value of y.

Training

- Exercise 1

Take a common factor of the following expression by grouping of terms:

ax + ay + bx + by

Solution

The first two are grouped, with the common factor "a" and the last two with the common factor "b":

ax + ay + bx + by = a (x + y) + b (x + y)

Once this is done, a new common factor is revealed, which is (x + y), so that:

ax + ay + bx + by = a (x + y) + b (x + y) = (x + y) (a + b)

Another way to group

This expression supports another way of grouping. Let's see what happens if the terms are rearranged and a group is made with those that contain x and another with those that contain y:

ax + ay + bx + by = ax + bx + ay + by = x (a + b) + y (a + b)

In this way the new common factor is (a + b):

ax + ay + bx + by = ax + bx + ay + by = x (a + b) + y (a + b) = (x + y) (a + b)

Which leads to the same result from the first grouping that was tested.

- Exercise 2

The following algebraic expression is required to be written as the product of two factors:

3rd³ - 3rd^twob + 9ab^two-to^two+ab-3b^two

Solution

This expression contains 6 terms. Let's try grouping first and fourth, second and third and finally fifth and sixth:

3rd³ - 3rd^twob + 9ab^two-to^two+ab-3b^two = (3a³ -to^two) + (- 3a^twob + 9ab^two) + (ab-3b^two)

Now each parenthesis is factored:

= (3a³ -to^two) + (- 3a^twob + 9ab^two) + (ab -3b^two) = a^two (3a - 1) + 3ab (3b -a) + b (a-3b)

At first glance it seems that the situation has been complicated, but the reader should not be discouraged, since we are going to rewrite the last term:

to^two (3a - 1) + 3ab (3b -a) + b (a-3b) = a^two (3a - 1) + 3ab (3b-a) - b (3b-a)

The last two terms now have a common factor, which is (3b-a), so they can be factored. It is very important not to lose sight of the first term^two (3a - 1), which must continue to accompany everything as an addition, even if you are not working with it:

to^two (3a - 1) + 3ab (3b-a) - b (3b-a) = a^two (3a - 1) + (3b-a) (3ab-b)

The expression has been reduced to two terms and a new common factor is discovered in the last one, which is "b". Now it remains:

to^two (3a - 1) + (3b-a) (3ab-b) = a^two (3a - 1) + b (3b-a) (3a-1)

The next common factor to appear is 3a - 1:

to^two (3a - 1) + b (3b-a) (3a-1) = (3a - 1) [a^two + b (3b-a)]

Or if you prefer without brackets:

(3a - 1) [a^two + b (3b-a)] = (3a - 1) (a^two -ab + 3b^two)

Can the reader find another way of grouping that leads to this same result??

References

1. Baldor, A. 1974. Elementary Algebra. Cultural Venezolana S.A.
2. Jiménez, R. 2008. Algebra. Prentice hall.
3. Main cases of factoring. Recovered from: julioprofe.net.
4. UNAM. Basic Mathematics: Factorization by grouping of terms. Faculty of Accounting and Administration.
5. Zill, D. 1984. Algebra and Trigonometry. MacGraw Hill.