This lesson introduces the
technique of factoring by grouping, as well as factoring the sum and
difference of cubes. Factoring by grouping builds on the ideas that
were presented in the section on factoring the common factor. While
there are many different types of grouping, as you will learn in higher
algebra courses, all of the grouping problems in this book involve four
terms, and they work by grouping the first two terms and the second two
terms together. If you do it right, a common factor will always
emerge! Remember, the method of grouping is one of trial and error. As
always, there is no substitute for practice and experience. The second
half of this lesson is the sum and difference of cubes, which may be
optional to your curriculum this first level of algebra. Remember, if
you can learn this topic now, it will help you later.
FACTORING
BY GROUPING
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. ax + bx
+ ay + by 6. ax
+ ac + bx + bc
7. ax −
bx + ay − by 8.
ax − ac + bx − bc
9. ax −
bx − ay + by 10.
ax − ac − bx + bc
= x (a −
b) − y ( a − b)
= ___( ) − ___( )
= (
)( ) =
( )( )
11. xy −
5x − 2y + 10
12. x2y +
xy2 − 5x
− 5y
=
= xy( ) −
5( )
=
=
13. x3
− x2
− 9x + 9
14. x3
− 5x2
− 4x + 20
= x2
( ) − 9( )
= x2
( ) − 4( )
= (
)( ) =
= (
)( )( )
=
15. x3
+ 7x2
− x − 7
16. x3
− 5x2
+ 25x − 125
= x2
( ) − 1(
) =
=
=
=
Does x2
+ 25 factor?
17. x3
+ 5x2
− 25x − 125
18. x3
− 5x2
− 25x + 125
19. x3
− 4x2 + 9x − 36
20. x3 + 4x2
− 9x − 36
21. x3
− 4x2 − 9x + 36
22. x3 − 9x2
− 4x + 36
23. x3
− 8x2 − x + 8
24. x3 −
8x2 + 4x − 32
SUM AND DIFFERENCE OF CUBES
(Optional−−Ask instructor!)
In this section, formulas and
procedures will allow you to factor expressions in the form x3
− y3 and also
x 3 + y
3. 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 +
xy + y2)
x3
+ y3 = (x + y)( x2 −
xy + 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."
The sum formula is
similar, in that the formula begins with the binomial that is
"the
first plus the second". Then use this binomial to
build the trinomial that follows, which except
for one sign is the same as the difference formula: take the "square of
the first" minus 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!
EXERCISES. Factor completely,
using the sum and difference of cubes formulas.
25. x3
− 8 26.
x3 − 125
= x3
− 23 [x=first;
2=second] = ( )3
− ( )3
= ( −
)(x2 + 2x
+ 22)
= ( − )( + + )
= (
)( ) = (
)( )
27. x3
− 64
28. x3
− 27
= ( )3
− ( )3
= ( )3
− ( )3
= (
)( ) = (
)( )
29. x3
+ 8 30. x3
+ 64
= x3
+ 23 [x=first;
2=second] = ( )3
+ ( )3
= ( + )(x2
− + )
= ( + )( − + )
= (
)( ) = (
)( )
31. x3
+ 125 32. x3
+ 27
= ( )3
+ ( )3
= ( )3
+ ( )3
= (
)( ) = ( )(
)
33. 8x3
− 125 34. 27x3
− 8y3
= (2x)3
− 53
[2x=first; 5=second] = ( )3
− ( )3
= ( −
)[(2x)2 + (2x)(5)
+ 52]
= ( − )[ + + ]
= (
)( ) = (
)( )
35. 64x3
+ 125 36. 27x3
+ 8y3
= ( )3
+ ( )3
= ( )3
+ ( )3
= (
)[ ] = (
)[ ]
= (
)( ) = (
)( )
37. 8x3
− 27y3
38. 125y3
− 8x3
39. 8x3
+ 1 40. 125y
3 − 1
In the next exercises, don't
forget to factor the common factor first.
41. 16x4
− 54x 42.
3x3
− 24y3
= 2x(8x3
− 27) =
= 2x[( 2x
)3 − ( 3 )3
] =
= 2x(___ −
___)(____+____+____) =
43. 5x4
+ 40x 44.
10x5y
+ 80x2y4
45. 3x5y5
− 81x2y2
46. 16x2y2
+ 250x2y5
ANSWERS 2.07
p. 174-180:
(NOTE: Factors may be given in any
order!)
1.
(y+7)(x+4);
2. (2y+5)(x+5); 3. (x+3)(x2+9);
4. (x-3)(x2+9); 5. (a+b)(x+y);
6. (x+c)(a+b);
7. (a-b)(x+y); 8. (x-c)(a+b);
9. (a-b)(x-y); 10. (x-c)(a-b);
11.
(y-5)(x-2); 12. (x+y)(xy-5);
13. (x-1)(x-3)(x+3);
14. (x-5)(x-2)(x+2);
15.
(x+7)(x-1)(x+1); 16. (x-5)(x2+25);
17. (x+5)2(x-5); 18. (x-5)2(x+5);
19. (x-4)(x2+9);
20.
(x+4(x-3)(x+3); 21. (x-4)(x-3)(x+3);
22. (x-9)(x-2)(x+2);
23. (x-8)(x-1)(x+1);
24.
(x-8)(x2+4); 25. (x-2)(x2+2x+4);
26. (x-5)(x2+5x+25);
27. (x-4)(x2+4x+16);
28.
(x-3)(x2+3x+9);
29. (x+2)(x2-2x+4);
30. (x+4)(x2-4x+16);
31. (x+5)(x2-5x+25);
32.
(x+3)(x2-3x+9);
33. (2x-5)(4x2+10x+25);
34. (3x-2y)(9x2+6xy+4y
2);
35.
(4x+5)(16x2-20x+25);
36. (3x+2y)(9x2-6xy+4y2);
37. (2x-3y)(4x2+6xy+9y2);
38.
(5y-2x)(25y2+10xy+4x2);
39. (2x+1)(4x2-2x+1);
40. (5y-1)(25y2+5y+1);
41.
2x(2x-3)(4x2+6x+9);
42. 3(x-2y) (x2+2xy+4y2);
43. 5x(x+2)(x2-2x+4);
44.
10x2y(x+2y)(x2-2xy+4y2);
45. 3x2y2(xy-3)(x2y2
+3xy+9); 46. 2x2y2(2+5y)(4-10y+25y2).
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