Factor By Grouping

Factor by grouping, in layman language, is simply defined as the grouping of terms with common factors before factoring a polynomial. A polynomial is a mathematical expression that is formed by variables and constants. The construction of such variables and constants is done by using operations of addition, subtraction, multiplication and non-negative whole number exponents (constants). Before we get down to factor by grouping methods, lets see some simple examples of polynomials.

Examples of Polynomials

• 1 – 4x
• 5x³ - 8
• -2.16x + 7x³ - 7/2
• 4.7x³ + 3.1x³ + 9x
• (x - 3)² + 6x + 1 = x² - 6x + 9 + 6x + 1 = (x² + 10)
• x² − 4x + 7

So what is not a polynomial?
x² − 4/x + 7xx^3/2 is an example of a term which is not a polynomial. The reason behind this being that the second term (4/x) incorporates division i.e., 4x^-1 and the third term (7x^3/2) has a fractional index. These two conditions fail to meet one of the criteria for an expression to be a polynomial. As I have mentioned earlier, a polynomial must have 'Non-negative whole number exponents'.

Factoring Polynomials By Grouping
Here are some examples you can refer to.

Example 1:
9a³ – 15a² + 3ba – 5b

Solution:
9a³ – 15a² + 3ba – 5b

(Step: 1)
= (9a³ – 15a²) + (3ba – 5b)

(Step: 2)
= 3a²(3a – 5) + b(3a – 5)

(Step: 3)
= (3a – 5)(3a² + b)

Steps:

• In the first step, terms with common factors have been grouped
• In the second, the greatest common factor (GCF) is removed
• Finally, in the third step, the distributive law [ a(b + c) = ab + ac ] has been used

In the above sum, the important point to consider is that (2a – 3) is the common factor.

Example 2:
pq + 4q – 2p – 8

Solution:
pq + 4q – 2p – 8

(Step: 1)
= pq – 2p + 4q – 8

(Step: 2)
= p(q – 2) + 4(q – 2)

(Step: 3)
= (p + 4) + (q – 2)

Steps:

• The first step was rearranging the terms of the polynomial so that two consecutive have a common factor [such as p]
• In the second step, common factors were derived from each of the two consecutive terms
• In the third step, the polynomial was factored by grouping the common factors

Example 3
a³ – ab² – a² + b³

Solution:
a³ – ab² – a²b + b³

(Step: 1)
= a³ – a²b – ab² + b³

(Step: 2)
= a²(a – b) – b²(a – b)

(Step: 3)
= (a²– b²) (a – b)

(Step: 4)
= (a – b) (a + b) (a – b)

(Step: 5)
= {(a + b) (a – b)²}

Steps:

• Here, from step 1 to step 3, same procedures have been followed as those in problem 2
• In step 4, (a²– b²) has been simplified to (a – b) (a + b)
• In step 5, (a – b) was the common factor, so it was grouped together for the final solution.

Factoring Trinomials By Grouping
In 'factor by grouping' in a trinomial, the 'split method is used'. Here's how it goes.

All trinomials of the form (ax² + bx + c) ['c' being a constant], whose leading co-efficient is not 1, can be simplified by factoring through the split method. First, find the product of a and c (a . c). Now, the factors of (a . c) must be added to get the center term 'b'. Put the factors with their appropriate signs (+ or –) in place of the center term. Finally, follow the method of grouping of common factors. Follow the example, to understand it better.

Example 4

4x² + 13x + 10                   [ ax² + bx + c]

Solution:
4x² + 13x + 10

(Step: 1)
= 4x² + 8x + 5x + 10

(Step: 2)
= 4x(x + 2) + 5(x + 2)

(Step: 3)
= (4x + 5 ) (x + 2)

Steps:

• In the expression, a = 4 and c = 10. So, a .c = 40. Now, factors of 40 which add up to 'b' (= 13) are 8 and 5. Thus, 8 + 5 = 13
• The factors with their appropriate signs are put in the first step
• In the second step, common factors were found out of each of the two consecutive terms
• Finally in the third step, common binomial (terms) were paired.

Example 5

8x² + 11x – 7                    [ ax² + bx + c]

Solution:
8x² + 11x – 7

(Step: 1)
= 8x² + 14x + (- 4x) – 7

(Step: 2)
= 8x² + 14x – 4x – 7

(Step: 3)
= 2x(4x + 7) – 1(4x + 7)

(Step: 4)
= (4x + 7) (2x – 1)

Steps:

• Similar to example 5, a = 8 & c = (- 7), So, a . c = (- 56). Factors of (- 56) which can add up to 11 (= b) are 14 and (- 4). So likewise, these factors were put with their respective signs
• In the second step, the expression was further simplified [14x + (- 4x) = 14x – 4x]
• In the third step, common factors from consecutive terms were derived
• In the last step, the trinomial expression was factored.

Likewise, there are several examples of factoring by grouping method. Books are the best source for study tips on problem solving and also for the basic and in-depth knowledge about this subject of factor by grouping. Regular practice with different kinds of polynomials will enhance your skills and help you to master this mathematical art of simplification. This should help with your homework and in enhancing your knowledge. Always remember that in a subject like 'Mathematics', nothing works better than 'loads' of practice!