Lesson Factoring on an Intermediate Level

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This Lesson (Factoring on an Intermediate Level) was created by by rapaljer(4551) About Me : View Source, Show
About rapaljer: Retired Professor of Mathematics from Seminole State College of Florida after 36 years.

This text was imported from http://www2.seminolestate.edu/rrapalje/IntermediateAlgebra/Int%20One%20Step%20Ch%202/2201_Factoring.htm by its author.

2.01  Factoring, Factoring, Factoring 

 from Intermediate Algebra: One Step at a Time © 1998-2011

 

p. 111–142

 

Dr. Robert J. Rapalje

Seminole State College of Florida

ANSWERS TO ALL EXERCISES ARE INCLUDED AT THE END OF THIS PAGE

 

To see selected solutions in Living C O L O R  click here!

See also :      Basic Algebra:  Factoring the Common Factor

PLEASE NOTE:      Links are now provided throughout this section for additional explanations!!

Of all the topics that you have studied in previous algebra courses, there is no topic more important, there is no topic that you need to review more than the topic of factoring.  Let's begin with a working definition of factoring.  What does it mean to factor something?  What would you say if you were asked to "factor the number 15"?  Without hesitation, you would probably answer    "3 times 5" or "5 times 3"!  The key word is "times."  When asked to factor a given number, you naturally answer with a product of two numbers. 

DEFINITION

To FACTOR means to EXPRESS AS A PRODUCT!

Factoring is an important skill that goes all the way back to your first algebra course, and it will continue to be a most important skill, especially in calculus.  There are many different types of factoring that become more complex as well as more abstract in the higher math.  Some of the exercises in this section will begin with a review of elementary factoring exercises, such as x2 – 9 or x2 – 6 x, and then progress into higher levels of  factoring.  Notice the increase in complexity and abstraction as you "grow" through this "one step at a time."

 

 GUIDELINES TO FACTORING

 

  1.  Common Factor (Factor Common Factor First!)

 

  2.  Trinomial (F OI L rearranged to spell F L OI)

 

  3.  Difference of Squares: X2 – Y2 = (X – Y)(X + Y)

 

      Diff of Cubes:         X3 – Y3 = (X – Y)(X2 + XY + Y2)

 

      Sum of Cubes:          X3 + Y3 = (X + Y)(X2 – XY + Y2)

 

  4.  Factoring by Grouping

 

 

FACTORING THE COMMON FACTOR

See also :   Basic Algebra:  Factoring the Common Factor

                                                         Intermediate Algebra:  Factoring the Common Factor in Living Color

     The first step in any factoring problem should be to try to “factor the common factor.”    Factoring the common factor is simply using the distributive property in reverse.


 

EXAMPLES DISTRIBUTIVE PROPERTY:            EXAMPLES OF FACTORING

           6(x + 7) = 6 x + 42                                                    6 x + 42 = 6(x + 7)

        7(2 x + 3) = 14 x + 21                                                14 x + 21 = 7(2 x + 3)

       9(3 x – 4y) = 27 x – 36y                                              27 x – 36y = 9(3 x – 4y)

      12(2 x + 1) = 24 x + 12                                                24 x + 12 = 12(2 x + 1)

5(3 x – 2y + 4) = 15 x – 10y + 20                              15 x – 10Y + 20 = 5(3 x – 2Y + 4)

      5 x (x + 4) = 5 x2  + 20 x                                           5 x2  + 20 x = 5 x (x + 4)   

When factoring the common factor, look for a number or variable that divides into both (or all) terms.  If there is more than one common factor, be sure to get the largest common factor you can find.  First write down the common factor.  Then, open parentheses, and put down all the other factors that are left over.  Again, follow the “one step” format of the following exercises.

 

EXERCISES: Factor completely.

 1.  5 x2 + 15 x                        2.  18 x + 24y                          3.  35 x y + 7 y

    = 5 x (                 )                 = 6(                 )                       = 7 y (                   )

 

 4.  5 x3 – 45 x2                        5.  16a + 24b – 8                    6.  12 x – 24 y + 48

    = 5 x 2(                  )              = 8(                   )                     = 12(                         )

 

 7.  x3 + 4 x2                            8.  4 y 3 + 8 y                          9.  16a3 – 24a2

 

 

10. 12z3 – 18z2                      11.  24 x2 + 12                        12.  16b2 + 48b3

 

 

 

13.    24 x3 + 24 x2                                             14.    16 x2 – 32 x3       

 

 

 

15.     8a + 12b – 20c                                          16.    40 x – 32 y + 64   

 

 

 

17.    42 y3 – 14 y2 + 49 y                                   18.   24 x3 + 24 x2 + 24 x 

 

 

 

19.    19 x3 + 19 x2 y + 38 x2                             20.     17 x3 – 34 x2

 

 

 

21.    y 5 – 14y3                                                   22.      x10 + 5 x3 

     = y 3(                   )                                               =  x3(                          )

 

From these examples, observe the rule listed below:

RULE

When factoring powers, take out the lowest power of the factor.  Then subtract exponents.

 

23.    y10 + 7 y4                         24.    x7 + 8 x5                           25.     16 x2 y3 – 12 x3 y2

                                                                                                           = 4 x2 y2(                               )

                                                       

26.    5 x5 y2 + 10 x4 y3            27.    8 x5 y3 + 12 x3 y4             28.       36 x3 y4 + 24 x2 y6

  

 

In each of the next exercises, observe how you move from the simple to the more complicated; from the concrete to the abstract.

29a)       y x + 7 x                                            30a)     4 x y + 3 y

           = x(                  )

    b)        ya + 7a                                                 b)    4 xa + 3a

           = a(                   )

 

    c)      y$ + 7$                                                   c)     4 x$ + 3$

           = $(                  )

 

    d)      y(Junk) + 7(Junk)                                  d)    4X(Junk) + 3(Junk)   

            = (Junk) (                   )

 

    e)     y(3 x+4) + 7(3 x+4)                                 e)    4 x(8 y–7) + 3(8 y–7)

            = (3 x+4) (                   )

 

31.    a(3 x+4) – 5(3 x+4)                                   32.    5q(8r+7) + 3(8r+7)

 

 

33.    5u(3 x+4) + 9v(3 x+4)                             34.     10x(8 y–7) – 3(8 y–7)

 

 

RULE

In order to factor a common factor, you must have an identical factor common to all terms.  Be sure to count terms first.

EXAMPLE:    Can you factor  5u(3 x+4) + 9v(3 x–4) in this manner?  

                         NO!  There is no factor common to both terms.

35.      x(x– y) – y (x– y)                                 36.     x(x– y) – y(x– y) + 4(x– y)

 

37.      x(x– y) + y(x– y) – 4(x– y)

 

 

38.      x(2 x+3 y) – y(2 x+3 y) + 4(2 x+3 y)

 

 

39.       (x + y)2 – z(x + y)                             40.    (x – y)2 – z(x – y)

        = (            )[(          ) – ___]

        = (            )(                    )

 

41.   (x – y)2 – y(x – y)                                  42.    (x + y)2 – y(x + y)

 

 

 

 

43.    (2 x + 3 y)2 – 5(2 x + 3 y)                  44.     (2 x – 3 y)2 – 5(2 x – 3 y)

 

 

 

 

45.   (2 x + 3 y)2 + 5(2 x + 3 y)                  46.     (x + y)2 + (x – y)( x + y)

 

 

 

 

47.  (3 x–2 y)2 – 2(3 x–2 y)( x–5 y)               48.     (5 x+3 y)2 – 4(5 x+3 y)( x+3 y)

 

 

 

 

TRINOMIALS

See also :   Basic Algebra:  Factoring Trinomials

                                        

 

     Do you remember the product of two binomials, F OI L, and the fact that the result is usually in a trinomial?  As examples, consider:

                Product Binomials      F       OI        L

                    (x + 2)( x + 5) =       x2   +  7x   +  10    

                    (x + 2)( x – 5) =       x2   –   3x  –  10    

                    (x – 2)( x + 5) =       x2   +   3x  –  10

                    (x – 2)( x – 5) =       x2   –   7x  +  10

     In each of these examples, you were given a product of two binomials, and with F OI L, in each case you obtained a trinomial with x2.  Now the problem will be to work these problems in reverse.  What if you were given a trinomial, such as x2+7 x+10 and asked to factor it––that is, to express it as a product.  This is the product of the two binomials.  When you factor 

the trinomial:                                         x2 + 7 x + 10 

you expect the product of binomials:   (          )(          ).

When factoring a trinomial, instead of thinking F OI L, you need to change the order and think F L OI In other words, you need to find the correct F (first times first) combination, then skip to the L (last times last).  Finally, check to make sure the OI (outer times outer, inner times inner) middle term is correct.

                                      F    OI    L

 x2 + 7 x + 10       Given trinomial to factor;     

(         )(          )      Product of two binomials;

(x       )( x       )      F term is x2, which is x times x;

(x       )(x       )     L term is +10.  Find two numbers whose product is +10.  Probably 2 times 5, or it could be    1 times 10.  Try 2 times 5.  Since the last sign is positive, it will be positive times positive, or negative times negative.  Also, the middle terms (O and I) must be added together.   

(x + 2)( x + 5)     OI term is 7x.  This means the outer times outer and the inner times inner terms must add up to 7x.    

[NOTICE: The order does not matter!  If you wrote (x+5)( x+2), this is exactly equivalent to (x + 2)( x + 5)WHY??]

     The following is a helpful summary of trinomial (F L OI) factoring.

    

RULES: 1.     When the sign of the LAST term is positive, the signs are the SAME. 

                         Find middle term by ADDING the O and I terms.

 

                2.      When the sign of the LAST term is negative, the signs are OPPOSITE. 

                          Find middle term by SUBTRACTING the O and I terms.

 

It is also worth noting that not all trinomials can be factored.  For examples, the trinomials x2 + x + 2, x2 + 2x + 6, and x2 – 4x – 6 cannot be factored.  These are called prime trinomials. 

 

 

EXERCISES.  Factor each of the following trinomials.

 1.     x2 + 3 x + 2                         2.       x2 + 9 x + 14                           3.     x2 + 7 x + 10

 

 

 4.     x2 –12 x + 35                       5.      x2 – 13 x + 22                          6.      x2 – 9 x + 18

 

 

 7.     x2 + 3 x – 40                         8.     x2 + 18 x – 40                          9.      x2 + 39 x – 40

 

 

10.     x2 – 15 x – 16                     11.      x2 – 6 x – 16                           12.      x2 –  x – 72

 

 

13.     x2 – 7 x – 18                       14.      x2 + 3 x – 18                          15.      x2 – 19 x + 60

 

 

16.     x2 – 6 x – 27                        17.     x2 +  2 x – 24                          18.      x2 – 5 x – 24

 

 

19.     x2 – 32 x + 60                     20.     x2 + 11 x – 60                        21.      x2 + 61 x + 60

 

 

22.     x2 – 23 x + 60                     23.      x2 – 17 x – 60                        24.      x2 – 59 x – 60

 

 

25.     x2 –  x – 56                          26.      x2 – x – 72                             27.      x2 + 17 x + 72

 

 

28.     x2 + 17 x + 42                    29.      x2 – 19 x – 42                        30.      x2 – 23 x + 42

 

 

     Frequently, it is necessary to FACTOR THE COMMON FACTOR FIRST (FCFF).  When there is a common factor in the problem, always remember to FCFF! (NOTE:  These exercises require TWO steps!)

 

EXERCISES:  Factor the trinomials completely.  Be sure to FCFF!!

31.     3x2 + 6 x – 9                        32.     5 x2 + 15 x – 20                    33.      8 x2 + 8 x – 48

          3(x2 + 2x – 3)                                5(                        )                           ___(                      )

         3(          )(          )                           5(             )(           )                         ___(           )(           )

 

34.     6 x2 + 6 x – 12                    35.     12 x2 + 36 x – 120                  36.       10 x2 + 30 x – 100

       ___(                       )

       ___(            )(             )       

 

37.      x3 – 4 x2 – 5 x                    38.       x3 + 5 x2 + 6 x                   39.       2x3 – 14 x2 + 20x

 

 

 

40.     5 x 3 + 5 x 2 – 10 x              41.      7 x 3 + 49 x 2 + 42 x              42.       8 x 3 – 40 x 2 + 32 x

 

 

 

43.     x 4x 3 – 20 x 2                44.      x 4 + 2 x 3 – 35 x 2                 45.      15 x 4 – 45 x 3 – 60 x 2

 

 

 

 

46.    20 x 4 – 20 x 3 – 120 x 2       47.      30 x 4 + 90 x 3 –300 x 2          48.      18 x 4 + 54 x3 + 36 x2

 

  

 

 

See also :  Basic Algebra :  Advanced Trinomial Factoring

                                                    

In the factoring of trinomials of the previous exercises, it may have been assumed that either the coefficient of x2 is 1 or that the coefficient can be factored as a common factor of the entire trinomial.  However, consider the example 5x2 + 6x + 1.  Remember that this should be factored by F L OI, where the F term must be 5x2, L must be 1, and the OI term must add up to 6x, as follows:

5 x2 + 6 x + 1

                                                                 (              )(               )

 Try the following examples (solutions are given below):

                    3 x 2 + 4 x + 1                           8 x 2 + 9 x + 1                             8 x 2 + 6 x + 1

                  (           )(            )                      (           )(            )                         (           )(            )

 

 

 

Solutions:      (3x + 1)(x + 1)                         (8 x + 1)( x + 1)                           (4 x + 1)(2 x + 1)

Of course, with larger numbers, with many more combinations of numbers this can become a very lengthy process of trial and error.  There are some systematic methods of factoring these trinomials, one of which will be presented at the end of the "Factoring by Grouping" section.  In problems that are not too difficult, the trial and error method will be fairly simple and more than adequate for now.

 

Consider the examples 5x2 + 8x + 3 and 5x2 + 16x + 3.   Again, these are factored by F L OI, where the F term must be 5x2, L must be 3, and the OI term must add up to 8x and 16x respectively, as follows: 

5x2 + 8 x + 3                  5x2 + 16 x + 3

                                 (            )(           )             (           )(            )

 Try the following examples (again solutions are given below):

                       3x2 + 10 x + 7                        3x2 + 22 x + 7                             3x2 + 4 x – 7

                     (           )(            )                    (            )(            )                       (            )(            )

 

 

 

Solutions:        (3 x + 7)( x + 1)                       (3 x + 1)( x + 7)                          (3 x + 7)( x – 1)

 

  REMEMBER:     1.      When the L term is positive add the O and I terms.

                                2.      When the L term is negative, subtract the O and I terms.

 EXERCISES:

 1.     3 x 2 + 4 x + 1                           2.     4 x 2 + 5 x + 1                            3.     7 x 2 – 8 x + 1

 

 

 4.     3 x 2 + 2 x – 1                            5.     4 x 2 + 3 x – 1                              6.     7 x 2 + 6 x – 1

 

 

 7.    10 x 2 – 7 x + 1                           8. 10 x 2 – 9 x – 1                                9.    10 x 2 + 3 x – 1

 

 

10.     3 x 2 + 8 x + 5                       11.  3x2 + 16x + 5                              12.    3 x 2 + 2 x – 5

 

 

13.     3 x 2 – 14 x – 5                       14.     3 x 2 – 34 x + 11                       15.    3 x 2 + 8 x – 11

 

 

16.     5 x 2 + 3 x – 8                        17.     5 x 2 + 39 x – 8                        18.     5 x 2 + 6 x – 8

 

 

19.     6 x 2 – 19 x + 8                       20.     6 x 2 – 13 x – 8                         21.     6 x 2 + 13 x – 8

 

 

22.    6 x 2 + 19 x + 10                      23.    6 x 2 + 11 x – 10                      24.     6 x 2 + 17 x + 10

 

  

See also :   Basic Algebra:  Factoring the Difference of Squares

                                                       

DIFFERENCE OF SQUARES; PERFECT SQUARE TRINOMIALS

                               EXAMPLES:                                                               GENERALIZATIONS:

                            x2 – 25 = (x – 5)( x + 5)                                                        x 2 – y2 = (x – y)( x + y)

                x2 + 10 x + 25 = (x + 5)( x + 5)                                            x2 + 2 x y + y2 = (x + y)( x + y)

                                        = (x + 5)2                                                                              = (x + y)2

                x2 – 10 x + 25 = (x – 5)( x – 5)                                             x2 – 2 x y + y2 = (x – y)( x – y)

                                       = (x – 5)2                                                                                = (x – y)2              

 1.     x 2 – 64                                               2.    x 2 –  100                                          3.      x2 – 81

   (x –____)( x +____)

 

 4.     x2 – 169                                             5.     x2 – a2                                              6.      x2 – b2

 

 

 7.     4 x 2  –   9                                          8.      64 x 2 – 121                                     9.      16 x 2 – 49

   (2 x –___)(2 x +___)      

  

10.       81 x2 – 25 y2                                 11.     49 x2  – 36 y2                                 12.      25 x2 – 144a2   

 

 

Don't forget that the first step in any factoring problem is to factor the common factor first.  This means that each of the following exercises will require two steps.  This is known as the "factoring two–step."

13.      9 x2 – 9                                           14.     3 x2 – 12                                        15.      5 x2 – 45

         9(                 )                                          3(                    )                                    ___(                  )

          9(          )(           )                                  3(          )(           )                                 ___(          )(          )

   

16.     4x2 – 64                                           17.      4x2 –  100                                     18.      8x2 – 72

  

 

 

 

19.     3x3 – 75x                                        20.       5x3 – 80x                                       21.      2x3 – 50x

 

 

 

 

22.      64 x4 – 4 x2y2                               23.      12 y4 – 12 x2 y2                              24.      79 y4 – 79 x2 y2

 

 

 

 

Factor each of the following perfect square trinomials:

25.       x 2 + 4 x + 4                                    26.       x 2 + 14 x + 49                           27.      x 2 + 20 x + 100

         =(          )(          )

        = (            )2

 

28.       x 2 – 12 x + 36                              29.       x 2 – 18 x + 81                                30.      x 2 – 24 x + 144

 

 

 

 

Remember to factor the common factor first:

31.      5 x 2 – 20 x + 20                            32.       2 x 2 – 20 x + 50                            33.      3 x 2 + 6 x + 3

 

 

 

 

34.      3x2 – 60 x + 300                             35.      6x2 – 36x + 54                               36.      4x2 + 32x + 64

 

 

 

 

37.      x 3 + 4x 2 + 4x                                  38.      x 3 – 12x2 + 36 x                          39.      9x3 – 18x2 + 9x

 

 

 

 

40.      6x 4 – 36x 3 + 54x 2                          41.     12y 4 – 48y 3 + 48y 2                      42.     2y 4 – 28y3 + 98y 2

 

 

 

 

 

In the next exercises, remember that sum of squares, such as x2  +  9 or x2  +  4, does not factor by this method.

43.       x 4 – 16                                                                44.        x 4 – 1           

  =( x – ____) (x + ____)

 =( x –___)( x +___)( x 2 + ___)

 

45.      x 4 – 81                                                                46.        x 4 – y 4           

 

 

 

47.      81 x 4 –  16 y 4                                                   48.       16 x 4 – 81

 

 

 

49.      x 4 + 10 x 2 + 9                                                50.        x 4 + 13 x 2 + 36    

 

 

 

51.      x 4 – 13 x 2 + 36                                               52.      x 4 – 29 x 2 + 100  

 

  

 

53.      y 4 + 5 y 2 – 36                                                 54.      y 4 – 18 y 2 + 81

 

 

 

55.      25 x 4 – 64 x 2                                                   56.       9 x 4 – 81 x 2

 

  

 

57.      9 x 4 – 36 x 2Y2                                                58.       9 x 3 – 54 x 2 + 81 x   

 

 

 

59.     8 x 3 + 80 x 2 + 200 x                                       60.      8 y 4 + 16 y 3 + 8 y 2

 

 

 

See also :   Intermediate Algebra:  Factoring Difference and Sum of Cubes in Living Color

SUM AND DIFFERENCE OF CUBES

In this section, formulas and procedures will enable you to factor expressions in the form x3 – y3 and also x3 + y3 Recall from previous sections that  x2 – y2 = (x – y)( x + y) and that x2 + y2 cannot be factored.  Begin with the multiplication problems:

           (X – Y)(X2 + XY + Y2)   =  X3 + X2Y + XY2

                                                                                 – X2Y – XY2 – Y3   

                                                   =   X3   –   Y3

 

           (X + Y)( X2 – XY + Y2 ) =  X3 – X2Y + XY2

                                                                                +  X2Y – XY2 + Y3

                                                   =  X3 + Y3 .

This derives the formulas, known as the “sum and difference of cubes formulas.”

 

              SUM AND DIFFERENCE OF CUBES FORMULAS

         x3 – y3 = (x – y)( x2 + x y + y2)

        x3 + y3 = (x + y)( x2 – x y + y2)

 

 

 

 

 

 

 

 

Translated into words, this means the sum or difference of two cubes can be factored into the product of a binomial times a trinomial.  Begin by taking the cube root of the perfect cubes.

In the difference formula, the binomial is “the first minus the second”.  Then use this binomial to “build” the trinomial that follows: take the “square of the first” plus the “product of the first and second” plus the “square of the second.” 

These formulas are really easy to remember.  Notice that the   x3 – y3 formula begins with (x – y).   The x3 + y3  formula begins with (x + y).  Next notice that the trinomial factor in both formulas is the same except for one sign.  This trinomial factor in each formula involves the “square of the first,” the “product of the two,” and the “square of the second.”  The first sign in the trinomial is the opposite of the sign of the binomial, and the last sign is always positive.  Finally, you will never be able to factor the resulting trinomial (by ordinary trinomial methods), so you need not even try (at this level).

Before getting into the exercises, be sure to be familiar with the perfect cubes: 13 = 1; 23 = 8; 33 = 27; 43 = 64; and 53 = 125.  Be able to recite these from memory: 1, 8, 27, 64, 125.  Practice them as you drive down the highway!

   

In each of the following, factor completely:

1.     x3 – 8                                                                     2.      x3  –  125

    =  x3 – 23 [x=first; 2=second]                                       = (         )3  –  (         )3

    = (      –     )( x 2 + 2 x + 22)                                         = (      –      )(       +        +        )

    = (              )(                           )                                    = (                )(                              )  

 

3.      x3  –  64                                                                 4.       x3  –  27

     = (       )3 – (       )3                                                      = (          )3 – (        )3

    = (               )(                            )                                  = (                )(                                )

 

5.      x3 + 8                                                                    6.      x3  +   64

     =  x3 + 23    [x=first; 2=second]                                    = (        )3 + (        )3

     = (      +       )( x2    –          +         )                               = (      +      )(         –         +          )

     = (              )(                                )                               = (               )(                                 )

 

 7.      x3 + 125                                                                 8.     x3 +  27

      = (        )3 + (        )3                                                       = (       )3 + (        )3

      = (            )(                                  )                                 = (             )(                                 )

                    

 9.      8 x3 – 125                                                              10.     27 x3  – 8 y3

     = (2 x)3 – 53 [2 x =first; 5=second]                                  = (         )3 – (         )3

     = (      –      )[(2 x)2 + (2 x)(5) + 52]                                 = (     –     )[       +       +        ]

     = (              )(                                 )                                 = (              )(                               )

 

11.     64 x3 + 125                                                            12.     27 x3 + 8 y3 

      = (           )3 + (          )3                                                   = (       )3 + (       )3

     = (              )[                                    ]                               = (               )[                                  ]

     = (                )(                                  )                               = (               )(                                 )

 

13.      8 x3 – 27 y3                                                             14.      125 y3  –  8 x3

 

 

 

15.      8 x3 + 1                                                                     16.     125 y3  –  1                           

 

 

 

In the next exercises, don’t forget the common factor first.

17.     16 x4 – 54 x                                                                   18.     3 x3 – 24 y3

     = 2x (8 x3 – 27)                    

     = 2x [( 2 x )3 – ( 3 )3 ]           

   = 2x ( ____ – ____ )( ____+_____+____ )  

 

19.    5 x4 + 40 x                                                                      20.     10 x5 y + 80 x2 y4

 

 

 

 

21.     3 x5 y5 – 81 x2 y2                                                          22.      16 x2 y2 + 250 x2 y5

 

 

 

 

23.      x6 – y9                                                                           24.     x6 + y9

     =  ( x )3 – ( y3 )3                                                                    = (           )3 + (           )3     

     = (       –       )[(       )2  +  (      )(     )+(      )2]                            = (                 ) [                                ]

     =      ___________________________                                 =     __________________________

 

25.     8 x6 + 125 y6                                                                 26.      8 x6 – 125 y6 

     =  (             )3 + (              )3     

     =

     =

 

Remember, if there is a common factor, factor this first.

27.    5 x7 + 40 x y9                                                               28.     5 x7 – 40 x y9 

    = 5 x(                       )

    = 5 x[(         )3 + (          )3 ]

    =

    =

 29.    16 x4 – 2x y6                                                                30.      25 x5 + 200 x8 y9

 

 

 

31.       x6  +  2x3  +  1                                                            32.     x6 – 9x3 + 8

     = (x3          )( x3           )

     =

     =

 33.       x6 – 7x3 – 8                                                                  34.       x6  –  64

                                                                                                      = (x –        )(x3  +        )

                                                                                                      =                        

 

See also :   Basic Algebra:  Factoring by Grouping

                                                         Intermediate Algebra:  Factoring by Grouping in Living Color

FACTORING BY GROUPING

This lesson introduces the technique of factoring by grouping.  It builds on the ideas that were presented in the section on factoring the common factor.  There are many different ways to group terms, some of which will be successful in the factoring process, others will not.  The method of grouping then is one of trial and error, with a few insights that may be helpful.  There is no substitute for practice and experience.

EXAMPLE 1: Factor  X3 + 2X2 + 8X + 16

SOLUTION: There are no common factors to all four terms.  It is not a trinomial, and nothing discussed so far works to factor this.  So, try grouping the first two terms together, and the last two terms together, and factor out the common factor within each grouping as follows:

                           (X3 + 2X2) + (8X + 16)

                        =  X2 (X + 2)+ 8(X + 2)

                       Notice that there is a common factor of (X+2) that can be factored out:

                       =  (X + 2)( X2 + 8)

 

EXAMPLE 2: Factor  XY – 4Y + 3X – 12

SOLUTION: Again, there are no common factors, and this is not a trinomial.  Group the first two and the last two terms together, and factor out the common factor from each grouping:      

                     (XY – 4Y) + (3X – 12) = Y(X – 4) + 3(X – 4)      Now, take out the common factor, which is (X– 4):

                                                              = (X – 4)(Y + 3)

EXAMPLE 3: Factor  XY – 4Y – 3X + 12

SOLUTION: Group the first two and the last two terms together, and factor out the common factor from each grouping:      

                        (XY – 4Y) + (–3X + 12) = Y(X – 4) + 3(–X + 4)                   

                    This time there is no common factor.  Try again, this time factoring a –3 from the last grouping.  This works!

                       XY – 4Y  – 3X + 12 = Y(X – 4) – 3(X – 4)

                                                        = (X – 4)(Y – 3)

EXAMPLE 4: Factor  XY – 4Y + 3X + 12

SOLUTION: Group the first two and the last two terms:   

                      XY – 4Y + 3X + 12 = Y(X – 4) + 3(X + 4)           

At this point, it is important to realize that no common factor resulted.  Do not try to factor out something that is not common to both groupings.  In fact, there is no way to group this problem to get a common factor.  This one cannot be factored by grouping.  In fact, it can’t be factored by any method.  Remember, not all problems can be factored.

Remember that the entire process of “grouping” is one of trial and error, and, as you will see later, there are different types of grouping. 

  

EXERCISES: Factor each of the following by grouping:

 1.       XY + 7X + 4Y + 28                                              2.     2XY + 5X + 10Y + 25

        = X(            ) + 4(            )                                        =

        = (                  )(                 )                                      =

 

3.       X3 + 3X2 + 9X + 27                                             4.     X3 – 3X2 + 9X – 27

      =                                                                                  =

      =                                                                                  =

 

5.      XY – 5X – 2Y + 10                                               6.     X2Y + XY2 – 5X – 5Y

      = X(            ) – 2(            )                                          = XY(             ) – 5(             )

      =                                                                                 =

 

 7.      X3 – X2 – 9X + 9                                                  8.     X3 – 5X2 – 4X + 20

       = X2 (          ) – 9(           )                                         = X2 (             ) – 4(             )

       = (             )  (               )                                           =

       = (             )(               )(              )                             =

 

 9.     X3 + 7X2 – X – 7                                                 10.     X3 – 5X2 + 25X – 125

      = X2 (           ) – 1(            )                                          =

      =                                                                                  = 

      =                                                                                 Does  X2 + 25 factor?

 

11.      X3 + 5X2 – 25X – 125                                       12.     X3 – 5X2 – 25X + 125

 

 

 

13.      X3 – 4X2 + 9X – 36                                          14.    X3 + 4X2 – 9X – 36        

 

 

 

15.      X3 – 4X2 – 9X + 36                                          16.     X3 – 9X2 – 4X + 36   

 

 

 

In the previous exercises, the first two terms and the last two terms were grouped in order to obtain a common factor.  As you might imagine, other groupings are frequently convenient.  As a prelude to such “other groupings,” the following exercises, which may be thought of as advanced trinomials or difference of squares, will help set the stage for what is to come.  Also, notice that you are moving to a higher level of abstraction in these exercises.

In the following exercises, notice the patterns that develop.

17a)  X2 – 9                                                                18a)  X2 – 64

 

     b)   π2 – 9                                                                   b)  π2 – 64

 

     c)    (Junk)2 – 9                                                         c) (Junk)2 – 64

       =[(          ) – 3][(          ) + 3]

       =(                )(                )

     d) (2X+3Y)2 – 9                                                         d) (2X+3Y)2 – 64

        =[(          ) – 3][(           ) + 3]

        =(                )(                 )

 

19a)  X2 – 5X + 6                                                        20a)  X2 – 5X – 6  

 

 

   b) (Junk)2 – 5(Junk) + 6                                               b) (Junk)2 – 5(Junk) – 6

    = [(         ) – 2][(          ) – 3]

    = (                  )(                   )

 

   c)  (2X+3Y)2 – 5(2X+3Y) + 6                                       c) (2X+3Y)2 – 5(2X+3Y) – 6

     = [(            ) – 2][(            ) – 3]

     =(                      )(                     )

21.    (2X–3Y)2 – 25                                                     22.    (2X–3Y)2 – 169

 

 

 

23.    (2X–3Y)2 – 4(2X–3Y) – 60                                24.    (2X–3Y)2 – 14(2X–3Y) + 40

 

 

  

25a)  X2 – 10X + 25                                                   26a)     X2 + 16X + 64   

  

  b) (Junk)2 – 10(Junk) + 25                                          b)    (Junk)2 + 16(Junk) + 64

 

  c) (2X+3Y)2 – 10(2X+3Y) + 25                                    c)    (2X+3Y)2 + 16(2X+3Y) + 64

 

 

27.    (2X–3Y)2 – 14(2X–3Y) + 49                              28.     (2X–3Y)2 + 10(2X–3Y) + 25

 

  

 

29.    (2X–3Y)2 + 20(2X–3Y) + 100                            30.    (2X–3Y)2 – 16(2X–3Y) + 64

 

 

 

EXAMPLE 5:  (X2 + 7X)2 + 16(X2 + 7X) + 60   This is a TRINOMIAL!  Remember F L OI ??

                          [                        ][                      ]    Product of two binomials

                          [(X2 + 7X)         ][(X2 + 7X)      ]     F = (X2 + 7X)2;    L = 60    

                         [(X2 + 7X)  +  10][(X2 +7X) +  6]     F = (X2 + 7X)2;     L =10·6;   OI = 16

                             (X2 + 7X + 10 )(X2 + 7X + 6)

                              (X + 5)(X + 2) (X + 6)(X + 1)

 

31.    (X2–2X)2 – 7(X2–2X) – 8                                   32.     (X2–5X)2+ 10(X2–5X) + 24      

     = [(X2–2X) –       ][(X2–2X) +        ]

     = ( X2 –2X –      )( X2 – 2X +       )

     =(           )(          )(           )(          )

or      (            )(            )(            )2

 

33.    (X2 –7X)2 + 16(X2 –7X) + 60                            34.      (X2 –5X)2– 2(X2 –5X) – 24      

 

 

 

 

35.     (X2–5X)2 – 36                                                   36.       (X2+5X)2 – 36

 

 

 

  

There are many different variations of factoring by grouping.  If there are four terms, sometimes it is "appropriate" to group the first two terms and the last two terms together.  Sometimes three of the terms "look good together," and it is appropriate to group three terms together, leaving the other term alone.  Sometimes there are more than four terms, and different groupings are “appropriate."  What makes a grouping "appropriate" is that the result is factorable.  In other words, it works!

37.       (X2 + 2XY + Y2)– 49                                      38.       X2 + 4XY + 4Y2 – 25

 

 

 

 

 

39.      X2 – 6XY + 9Y2 – 36                                       40.       X2 – 8XY + 16Y2 – 81

 

 

 

 

 

41.      25 – X2 + 2XY – Y2                                        42.       64 – X2 + 4XY – 4Y2

       = 25 – (X2 – 2XY + Y2)                                               = 64 – (                             )

      = 25 – (X – Y)2                                                             = 64 – (                  )2 

      = [5 – (            )][5 + (           )]                                     =  [         –  (              )][        + (               )]

      = (                    )(                     )                                    = (                               )(                             )

 

43.    16 – X2 – 10XY – 25Y2                                       44.       9 – X2 – 4XY – 4Y2

 

  

 

45.     X2 + 2XY + Y2   +   7X + 7Y   + 10                  46.      X2 – 2XY + Y2  +   5X – 5Y   +   6