ALGEBRA
Sec. 02
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Márquez
FACTOR POLYNOMIALS by GROUPING
THE IDEA FACTOR POLYNOMIALS by GROUPING
This is a very powerful method. Despite this, it may sometimes fail. It is not always clear if this method will work
or not on a particular problem. You must try it and practice. If you can’t factor a polynomial using this method,
relax. Later we will learn a few more ideas to factor.
Examples
1. 8 + 8y + 6y 3 + 6y 4
8 + 8y + 6y 3 + 6y 4 = (8 · 1 + 8y) + (6y 3 · 1 + 6y 3 · y)
(ALA and M.Id)
3
= 8(1 + y) + 6y (1 + y)
(D.L.)
3
= (8 + 6y )(1 + y)
(D.L.)
3
= 2(4 + 3y )(1 + y)
(D.L., BI)
note we could, in this case, have used the DL METHOD first, then use the grouping method.
2. 1 + 3x + x2 + 3x3
1 + 3x + x2 + 3x3 = (1 + 3x) + (1 · x2 + 3x · x2 )
(ALA and M.Id)
2
= (1 + 3x) · 1 + (1 + 3x)x
(MId, D.L.)
2
= (1 + 3x)(1 + x )
(D.L.)
3. 1 + 3x + x2 + 3x3
Here, we factor the same polynomial as above, but we group not the first two and the last terms, but the first
with the third term and we group the second with the fourth, , just for fun. Aside, from fun, this points out
that sometimes various arrangements of groupings are effective.
1 + 3x + x2 + 3x3 = 1 + x2 + 3x + 3x3
2
(CoLA, BI)
3
= (1 + x ) + (3x + 3x )
2
(ALA)
2
= 1 · (1 + x ) + 3x(1 + x )
2
= (1 + 3x)(1 + x )
(MiD, BI, DL)
(BI, DL)
4. 1 + 3x − x2 − 3x3
1 + 3x − x2 − 3x3 = (1 + 3x) + (−x2 + −3x3 )
(ALA)
2
2
= (1 + 3x) + (1 · −x + 3x · −x )
2
= (1 + 3x) + (1 + 3x)(−x )
(D.L.)
2
= (1 + 3x) · 1 + (1 + 3x)(−x )
2
= (1 + 3x)(1 + (−x ))
2
= (1 + 3x)(1 + −x )
(MId, N-Expo, CLM)
(M.Id)
(D.L.)
(D.L.(to be continued...))
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2007-2008
MathHands.com
ALGEBRA
Sec. 02
MathHands.com
Márquez
1. If possible factor by grouping, if not, state so.
2x3 + x2 + 2x + 1
Solution:
2x3 + x2 + 2x + 1
= 2x3 + x2 + 2x + 1
= (2x + 1) x2 + (2x + 1) 1
= (2x + 1) x2 + 1
(given)
(ALA)
(DL, BI, note: looks like someone distributed (2x + 1))
(DL)
2. If possible factor by grouping, if not, state so.
x3 + x2 + x + 1
Solution:
x3 + x2 + x + 1
= x3 + x2 + x + 1
= (x + 1) x2 + (x + 1) 1
= (x + 1) x2 + 1
(given)
(ALA)
(DL, BI, note: looks like someone distributed (x + 1))
(DL)
3. If possible factor by grouping, if not, state so.
x4 + x3 + x + 1
Solution:
x4 + x3 + x + 1
= x4 + x3 + x + 1
= (x + 1) x3 + (x + 1) 1
= (x + 1) x3 + 1
(given)
(ALA)
(DL, BI, note: looks like someone distributed (x + 1))
(DL)
4. If possible factor by grouping, if not, state so.
2x4 + x3 + 2x + 1
c
2007-2008
MathHands.com
ALGEBRA
Sec. 02
MathHands.com
Márquez
Solution:
2x4 + x3 + 2x + 1
= 2x4 + x3 + 2x + 1
= (2x + 1) x3 + (2x + 1) 1
= (2x + 1) x3 + 1
(given)
(ALA)
(DL, BI, note: looks like someone distributed (2x + 1))
(DL)
5. If possible factor by grouping, if not, state so.
4x3 + x2 + 8x + 2
Solution:
4x3 + x2 + 8x + 2
= 4x3 + x2 + 8x + 2
= (4x + 1) x2 + (4x + 1) 2
= (4x + 1) x2 + 2
(given)
(ALA)
(DL, BI, note: looks like someone distributed (4x + 1))
(DL)
6. If possible factor by grouping, if not, state so.
− 5x6 + x5 + 10x − 2
Solution:
− 5x6 + x5 + 10x − 2
= − 5x6 + x5 + 10x − 2
= ( − 5x + 1) x5 + ( − 5x + 1) − 2
= ( − 5x + 1) x5 + − 2
(given)
(ALA)
(DL, BI, note: looks like someone distributed ( − 5x + 1))
(DL)
7. If possible factor by grouping, if not, state so.
x2 − 3x + − 3x + 9
Solution:
x2 − 3x + − 3x + 9
= x2 − 3x + − 3x + 9
= (x − 3) x + (x − 3) − 3
= (x − 3) x + − 3
(given)
(ALA)
(DL, BI, note: looks like someone distributed (x − 3))
(DL)
c
2007-2008
MathHands.com
ALGEBRA
Sec. 02
MathHands.com
Márquez
8. If possible factor by grouping, if not, state so.
x8 − 4x7 + − x + 4
Solution:
x8 − 4x7 + − x + 4
= x8 − 4x7 + − x + 4
= (x − 4) x7 + (x − 4) − 1
= (x − 4) x7 + − 1
(given)
(ALA)
(DL, BI, note: looks like someone distributed (x − 4))
(DL)
9. If possible factor by grouping, if not, state so.
x5 + x3 + x2 + 1
Solution:
x5 + x3 + x2 + 1
= x5 + x3 + x2 + 1
= (x2 + 1) x3 + (x2 + 1) 1
= (x2 + 1) x3 + 1
(given)
(ALA)
2
(DL, BI, note: looks like someone distributed (x + 1))
(DL)
10. If possible factor by grouping, if not, state so.
x5 + x4 + x3 + x2 + x + 1
Solution:
x5 + x4 + x3 + x2 + x + 1
= x5 + x4 + x3 + x2 + x + 1
= (x2 + x + 1) x3 + (x2 + x + 1) 1
= (x2 + x + 1) x3 + 1
(given)
(ALA)
2
(DL, BI, note: looks like someone distributed (x + x + 1))
(DL)
11. If possible factor by grouping, if not, state so.
9x3 − 6x2 + 15x + − 3x2 + 2x − 5
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2007-2008
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ALGEBRA
Sec. 02
MathHands.com
Márquez
Solution:
9x3 − 6x2 + 15x + − 3x2 + 2x − 5
(given)
3
2
2
= 9x − 6x + 15x + − 3x + 2x − 5
(ALA)
2
2
= (3x − 2x + 5) 3x + (3x − 2x + 5) − 1
(DL, BI, note: looks like someone distributed (3x2 − 2x + 5))
(DL)
= (3x2 − 2x + 5) 3x + − 1
12. If possible factor by grouping, if not, state so.
2x3 − x2 + − 6x + 3
Solution:
2x3 − x2 + − 6x + 3
= 2x3 − x2 + − 6x + 3
= (2x − 1) x2 + (2x − 1) − 3
= (2x − 1) x2 + − 3
(given)
(ALA)
(DL, BI, note: looks like someone distributed (2x − 1))
(DL)
13. If possible factor by grouping, if not, state so.
3x7 − x3 + − 9x4 + 3
Solution:
3x7 − x3 + − 9x4 + 3
= 3x7 − x3 + − 9x4 + 3
= (3x4 − 1) x3 + (3x4 − 1) − 3
= (3x4 − 1) x3 + − 3
(given)
(ALA)
4
(DL, BI, note: looks like someone distributed (3x − 1))
(DL)
14. If possible factor by grouping, if not, state so.
3y 3 x2 + y 3 + − 3x3 − x
c
2007-2008
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ALGEBRA
Sec. 02
MathHands.com
Márquez
Solution:
3y 3 x2 + y 3 + − 3x3 − x
= 3y 3 x2 + y 3 + − 3x3 − x
= (3x2 + 1) y 3 + (3x2 + 1) − x
= (3x2 + 1) y 3 + − x
(given)
(ALA)
2
(DL, BI, note: looks like someone distributed (3x + 1))
(DL)
15. Factor by grouping if possible.
x2 + x + 1
Solution: as is... the grouping method alone fails....
16. If possible factor by grouping, if not, state so.
3y 3 x2 + y 3 + − 3x3 − x
Solution:
3y 3 x2 + y 3 + − 3x3 − x
= 3y 3 x2 + y 3 + − 3x3 − x
= (3x2 + 1) y 3 + (3x2 + 1) − x
= (3x2 + 1) y 3 + − x
(given)
(ALA)
2
(DL, BI, note: looks like someone distributed (3x + 1))
(DL)
17. If possible factor by grouping, if not, state so.
2
2
x+2
+ −
x−2
Solution:
2
2
x+2
+ −
!
2
2
=
x+2
2
= (x + 2)
"
2
= (x + 2)
"
#
x−2
+
−
+ (x + 2) −
+ −
#
(given)
x−2
(ALA)
(DL, BI, note: looks like someone distributed (x + 2))
(DL)
c
2007-2008
MathHands.com
ALGEBRA
Sec. 02
MathHands.com
Márquez
18. If possible factor by grouping, if not, state so.
2
2
x+2
+ −
x−2
Solution:
2
2
x+2

=
x+2

2

2
= (x + 2) 
= (x + 2) 
x−2
 +
−
+ −

2
2

 + (x + 2) −
+ −
(given)
x−2
(ALA)
(DL, BI, note: looks like someone distributed (x + 2))

(DL)

19. The Following can be factored using the grouping method.
9x3 + 15x + − 3x2 − 5
A. TRUE
B. FALSE
20. The Following can be factored using the grouping method.
9x2 + 15x + − 4x − 5
A. TRUE
B. FALSE
c
2007-2008
MathHands.com