Factor By Grouping
Having a tough time in learning the procedures of factor by grouping? Here are some simple steps you can follow and practice.

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
x2 − 4/x + 7xx3/2 is an example of a term which is not a polynomial. This is because, the second term (4/x) incorporates division i.e., 4x-1 and the third term (7x3/2) has a fractional index. These two conditions fail to meet one of the criterias 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 applied
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 and grouping the terms of the polynomial so that the first two terms 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 factorized
a3 - ab2 - a2b + 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.
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)
= (4x2 + 8x) + (5x + 10)
(Step: 3)
= 4x(x + 2) + 5(x + 2)
(Step: 4)
= (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, factors were grouped
- In the third step, common factors were found out of each of the two consecutive terms
- Finally in the fourth step, common binomial (terms) were paired.
8x2 + 10x - 7 [ ax2 + bx + c]
Solution:
8x2 + 10x - 7
(Step: 1)
= 8x2 + 14x + (- 4x) - 7
(Step: 2)
= 8x2 + 14x - 4x - 7
(Step: 3)
= (8x2 + 14x) - (4x + 7)
(Step: 4)
= 2x(4x + 7) - 1(4x + 7)
(Step: 5)
= (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 10 (= 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, factors were grouped
- In the fourth step, common factors from consecutive terms were derived
- In the last step, the trinomial expression was factored.
Example #6
2x3 - 8x2 - 9x + 36
Solution:
(2x3 - 8x2) - (9x - 36)
= 2x2 (x - 4) - 9 (x - 4)
= (x - 4) (2x2 - 9)
Example #7
x3 - 3x2 + 10x - 30
Solution:
(x3 - 3x2) + (10x - 30)
= x2 (x - 3) + 10 (x - 3)
= (x - 3) (x2 + 10)
Example #8
2a2b + 30ab + 112b
Solution:
2a2b + 30ab + 112b
= 2b (a2 + 15a + 56)
= 2b (a2 + 8a + 7a + 56)
= 2b {a (a + 8) + 7 (a + 8)}
= 2b {(a + 8) (a + 7)}
Likewise, there are several examples of factoring by grouping method. Books are the best source for study tips on problem solving, and for the basic and in-depth knowledge about this subject of factor by grouping. Regular practice with different types of polynomials will enhance your skills and help you to master this mathematical art of simplification. Always remember that in a subject like 'Mathematics', nothing works better than 'loads' of practice!
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