por He common factor by grouping of terms is an algebraic procedure that allows us to write some algebraic expressions in the form of factors. In order to achieve this objective, one must first conveniently group the expression 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 it is mastered. Let’s first look at an illustrative example described step by step. Then the reader can apply what has been learned in each of the exercises that will appear later.
For example, suppose you need to factor the following expression:
2×2 + 2xy – 3zx – 3zy
This algebraic expression consists of 4 monomials or terms, separated by signs + and -, namely:
2×2, 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 factor common 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 does appear that helps to write the expression as the product of two or more factors.
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examples
Factor the expression: 2×2 + 2xy – 3zx – 3zy
Step 1: Group
2×2 + 2xy – 3zx – 3zy = (2×2 + 2xy) + (-3zx – 3zy)
Step 2: Take the common factor of each group
2×2 + 2xy – 3zx – 3zy =
=(2×2 + 2xy) – (3zx + 3zy) =
=2x(x+y) – 3z(x+y)
Yoimportant: the negative sign is also a common factor that must be taken into account.
Now notice that the parenthesis (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
2×2 + 2xy – 3zx – 3zy = (x+y) (2x – 3z)
With the previous result, the goal of factorization 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) and (2x – 3z).
Important points about the common factor by grouping
question 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 coincide with the original, if not, there is an error.
In the example above, the result is worked in reverse, to check that it is correct:
(x+y) (2x – 3z) = 2×2 -3zx +2xy – 3zy
Since 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 differently?
Answer: There are algebraic expressions that admit 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:
2×2 + 2xy – 3zx – 3zy = (2×2 – 3zx) + (2xy – 3zy)
And you can check that the result is the same as what 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 facilitates the calculations. For example, suppose you want to make 2×2 + 2xy – 3zx – 3zy equal to 0. What are 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 equal to 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.
Exercises
– Exercise 1
Take a common factor from the following expression by grouping 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 as the first form of grouping that was tried.
– Exercise 2
It is required to write the following algebraic expression as the product of two factors:
3a3 – 3a2b+9ab2-a2+ab-3b2
Solution
This expression contains 6 terms. Let’s try grouping first and fourth, second and third and finally fifth and sixth:
3a3 – 3a2b+9ab2-a2+ab-3b2 = (3a3 -a2) + (– 3a2b+9ab2) + (ab-3b2)
Now factor each parenthesis:
= (3a3 -a2) + (– 3a2b+9ab2) + (ab -3b2) = a2 (3a – 1) + 3ab(3b –a ) + b(a-3b)
At first glance it seems that the situation has become complicated, but the reader should not be discouraged, since we are going to rewrite the last term:
a2 (3a – 1) + 3ab(3b –a ) + b(a-3b) = a2 (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 a2 (3a – 1), which must continue to accompany everything as an addition, even if you are not working with it:
a2 (3a – 1) + 3ab (3b-a) – b(3b-a) = a2 (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:
a2 (3a – 1) + (3b-a)(3ab-b) = a2 (3a – 1) +b (3b-a)(3a-1)
The next common factor to appear is 3a – 1:
a2 (3a – 1) +b (3b-a)(3a-1) = (3a – 1)[ a2 + b(3b-a)]
Or if you prefer without brackets:
(3a – 1)[ a2 + b(3b-a)] = (3a – 1)(a2 –ab + 3b2)
Can the reader find another way of grouping that leads to this same result?
References
Baldor, A. 1974. Elementary Algebra. Cultural Venezolana SA Jiménez, R. 2008. Algebra. Prentice Hall. Main cases of factorization. Retrieved from: julioprofe.net. UNAM. Basic Mathematics: Factorization by grouping of terms. Faculty of Accounting and Administration. Zill, D. 1984. Algebra and Trigonometry. MacGraw Hill.