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Factoring by Grouping

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary: This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Factoring is an essential skill for success in algebra and higher level mathematics courses. Therefore, we have taken great care in developing the student's understanding of the factorization process. The technique is consistently illustrated by displaying an empty set of parentheses and describing the thought process used to discover the terms that are to be placed inside the parentheses. The factoring scheme for special products is presented with both verbal and symbolic descriptions, since not all students can interpret symbolic descriptions alone. Two techniques, the standard "trial and error" method, and the "collect and discard" method (a method similar to the "ac" method), are presented for factoring trinomials with leading coefficients different from 1. Objectives of this module: know how to factor a polynomial using the grouping method and when to try the grouping method.

Overview

  • Using Grouping to Factor a Polynomial
  • Knowing when to Try the Grouping Method

Using Grouping to Factor a Polynomial

Sometimes a polynomial will not have a particular factor common to every term. However, we may still be able to produce a factored form for the polynomial.

The polynomial x3+3 x26 x18 x 3 3 x 2 6 x 18 has no single factor that is common to every term. However, we notice that if we group together the first two terms and the second two terms, we see that each resulting binomial has a particular factor common to both terms.

The polynomial 'x cubed plus three x squared minus six x minus eighteen'. The first two terms of the polynomial have x square in common, and the last two terms of the polynomial have negative six in common.

Factor x 2 x 2 out of the first two terms, and factor 6 6 out of the second two terms.

x 2 (x+3)6(x+3) x 2 (x+3)6(x+3)

Now look closely at this binomial. Each of the two terms contains the factor (x+3) (x+3) .

Factor out (x+3) (x+3) .
(x+3)( x 2 6) (x+3)( x 2 6) is the final factorization.

x 3 +3 x 2 6x18=(x+3)( x 2 6) x 3 +3 x 2 6x18=(x+3)( x 2 6)

Knowing when to Try the Grouping Method

We are alerted to the idea of grouping when the polynomial we are considering has either of these qualities:

  1. no factor common to all terms
  2. an even number of terms

When factoring by grouping, the sign (+or) (+or) of the factor we are taking out will usually (but not always) be the same as the sign of the first term in that group.

Sample Set A

Example 1

Factor 8 a 2 b 4 4 b 4 +14 a 2 7 8 a 2 b 4 4 b 4 +14 a 2 7 .

  1. We notice there is no factor common to all terms.
  2. We see there are four terms, an even number.
  3. We see that terms 1 and 2 have +4 b 4 +4 b 4 in common (since the 1st term in the group is +8 a 2 b 4 ) +8 a 2 b 4 ) .
  4. We notice that the 3rd and 4th terms have +7 +7 in common (since the 1st term in the group is +14 a 2 +14 a 2 ).

    The equation eight a squared b to the fourth power minus four b to the fourth power plus fourteen a squared minus seven equals the sum of the product of four b to the fourth power and two a square minus one, and the product of seven and two a square minus 1. The two terms on the right side have two a square minus one in common. 8 a 2 b 4 4 b 4 +14 a 2 7= (2a 2 -1)(4b 4 +7) 8 a 2 b 4 4 b 4 +14 a 2 7= (2a 2 -1)(4b 4 +7)

Practice Set A

Use the grouping method to factor the following polynomials.

Exercise 1

a x+a y+b x+b y a x a y b x b y

Solution

(a+b)(x+y) (a+b)(x+y)

Exercise 2

2am+8m+5an+20n 2am+8m+5an+20n

Solution

(2m+5n)(a+4) (2m+5n)(a+4)

Exercise 3

a 2 x 3 +4 a 2 y 3 +3b x 3 +12b y 3 a 2 x 3 +4 a 2 y 3 +3b x 3 +12b y 3

Solution

( a 2 +3b)( x 3 +4 y 3 ) ( a 2 +3b)( x 3 +4 y 3 )

Exercise 4

15mx+10nx6my4ny 15mx+10nx6my4ny

Solution

(5x2y)(3m+2n) (5x2y)(3m+2n)

Exercise 5

40abx24abxy35 c 2 x+21 c 2 xy 40abx24abxy35 c 2 x+21 c 2 xy

Solution

x(8ab7 c 2 )(53y) x(8ab7 c 2 )(53y)

Exercise 6

When factoring the polynomial 8 a 2 b 4 4 b 4 + 14 a 2 7 8 a 2 b 4 4 b 4 14 a 2 7 in Sample Set A, we grouped together terms1 and 2 and 3 and 4. Could we have grouped together terms1 and 3 and 2 and 4? Try this.
8 a 2 b 4 4 b 4 +14 a 2 7= 8 a 2 b 4 4 b 4 +14 a 2 7=

Solution

yes

Do we get the same result? If the results do not look precisely the same, recall the commutative property of multiplication.

Exercises

For the following problems, use the grouping method to factor the polynomials. Some polynomials may not be factorable using the grouping method.

Exercise 7

2ab+3a+18b+27 2ab+3a+18b+27

Solution

( 2b+3 )( a+9 ) ( 2b+3 )( a+9 )

Exercise 8

xy7x+4y28 xy7x+4y28

Exercise 9

xy+x+3y+3 xy+x+3y+3

Solution

( y+1 )( x+3 ) ( y+1 )( x+3 )

Exercise 10

mp+3mq+np+3nq mp+3mq+np+3nq

Exercise 11

ar+4as+5br+20bs ar+4as+5br+20bs

Solution

( a+5b )( r+4s ) ( a+5b )( r+4s )

Exercise 12

14ax6bx+21ay9by 14ax6bx+21ay9by

Exercise 13

12mx6bx+21ay9by 12mx6bx+21ay9by

Solution

3( 4mx2bx+7ay3by ) 3( 4mx2bx+7ay3by )  Not factorable by grouping

Exercise 14

36ak8ah27bk+6bh 36ak8ah27bk+6bh

Exercise 15

a 2 b 2 +2 a 2 +3 b 2 +6 a 2 b 2 +2 a 2 +3 b 2 +6

Solution

( a 2 +3 )( b 2 +2 ) ( a 2 +3 )( b 2 +2 )

Exercise 16

3 n 2 +6n+9 m 3 +12m 3 n 2 +6n+9 m 3 +12m

Exercise 17

8 y 4 5 y 3 +12 z 2 10z 8 y 4 5 y 3 +12 z 2 10z

Solution

Not factorable by grouping

Exercise 18

x 2 +4x3 y 2 +y x 2 +4x3 y 2 +y

Exercise 19

x 2 3x+xy3y x 2 3x+xy3y

Solution

( x+y )( x3 ) ( x+y )( x3 )

Exercise 20

2 n 2 +12n5mn30m 2 n 2 +12n5mn30m

Exercise 21

4pq7p+3 q 2 21 4pq7p+3 q 2 21

Solution

Not factorable by grouping

Exercise 22

8 x 2 +16xy5x10y 8 x 2 +16xy5x10y

Exercise 23

12 s 2 27s8st+18t 12 s 2 27s8st+18t

Solution

( 4s9 )( 3s2t ) ( 4s9 )( 3s2t )

Exercise 24

15 x 2 12x10xy+8y 15 x 2 12x10xy+8y

Exercise 25

a 4 b 4 +3 a 5 b 5 +2 a 2 b 2 +6 a 3 b 3 a 4 b 4 +3 a 5 b 5 +2 a 2 b 2 +6 a 3 b 3

Solution

a 2 b 2 ( a 2 b 2 +2 )( 1+3ab ) a 2 b 2 ( a 2 b 2 +2 )( 1+3ab )

Exercise 26

4 a 3 bc14 a 2 b c 3 +10ab c 2 35b c 4 4 a 3 bc14 a 2 b c 3 +10ab c 2 35b c 4

Exercise 27

5 x 2 y 3 z+3 x 3 yw10 y 3 z 2 6wxyz 5 x 2 y 3 z+3 x 3 yw10 y 3 z 2 6wxyz

Solution

y( 5 y 2 z+3xw )( x 2 2z ) y( 5 y 2 z+3xw )( x 2 2z )

Exercise 28

a 3 b 2 cd+ab c 2 dx a 2 bxyc x 2 y a 3 b 2 cd+ab c 2 dx a 2 bxyc x 2 y

Exercise 29

5 m 10 n 17 p 3 m 6 n 7 p 4 40 m 4 n 10 q t 2 +8pq t 2 5 m 10 n 17 p 3 m 6 n 7 p 4 40 m 4 n 10 q t 2 +8pq t 2

Solution

( m 6 n 7 p 3 8q t 2 )( 5 m 4 n 10 p ) ( m 6 n 7 p 3 8q t 2 )( 5 m 4 n 10 p )

Exercises for Review

Exercise 30

((Reference)) Simplify ( x 5 y 3 )( x 2 y) ( x 5 y 3 )( x 2 y) .

Exercise 31

((Reference)) Use scientific notation to find the product of (3× 10 5 )(2× 10 2 ) (3× 10 5 )(2× 10 2 ) .

Solution

6× 10 3 6× 10 3

Exercise 32

((Reference)) Find the domain of the equation y= 6 x+5 y= 6 x+5 .

Exercise 33

((Reference)) Construct the graph of the inequality y2 y2 .

A horizontal line with arrows on both ends.

Solution

A number line with arrows on each end, labeled from negative three to three in increments of one. There is a closed circle at negative two. A dark arrow is originating from this circle, and heading towrads the right of negative two.

Exercise 34

((Reference)) Factor 8 a 4 b 4 +12 a 3 b 5 8 a 2 b 3 8 a 4 b 4 +12 a 3 b 5 8 a 2 b 3 .

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