Techniques for Factoring TrinomialsExercise Set A.1: Factoring Polynomials 708 University of Houston Department of Mathematics 48. 4 28xx2 49. 2 18x2 50. 88x2 51. 5 20xx42 52. 3 75xx3
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APPENDICES
University of Houston Department of Mathematics 696
Appendices
Appendix A.1: Factoring Polynomials
Techniques for Factoring Trinomials
Techniques for Factoring Trinomials
Factorability Test for Trinomials:
Example:
Solution:
APPENDIX A.1 Factoring Polynomials
MATH 1330 Precalculus 697
Factoring Trinomials with Leading Coefficient 1:
APPENDICES
University of Houston Department of Mathematics 698
APPENDIX A.1 Factoring Polynomials
MATH 1330 Precalculus 699
Example:
Solution:
APPENDICES
University of Houston Department of Mathematics 700
Factoring Trinomials with Leading Coefficient Different from 1:
APPENDIX A.1 Factoring Polynomials
MATH 1330 Precalculus 701
APPENDICES
University of Houston Department of Mathematics 702
Example:
Solution:
Additional Example 1:
(a) 22 3 8x x
(b) 242 25 3x x
APPENDIX A.1 Factoring Polynomials
MATH 1330 Precalculus 703
Solution:
Additional Example 2:
Solution:
APPENDICES
University of Houston Department of Mathematics 704
Additional Example 3:
Solution:
APPENDIX A.1 Factoring Polynomials
MATH 1330 Precalculus 705
Additional Example 4:
Solution:
APPENDICES
University of Houston Department of Mathematics 706
Additional Example 5:
Solution:
Exercise Set A.1: Factoring Polynomials
MATH 1330 Precalculus 707
At times, it can be difficult to tell whether or not a
quadratic of the form 2
ax bx c can be factored
into the form dx e fx g , where a, b, c, d, e, f,
and g are integers. If 2
4b ac is a perfect square, then
the quadratic can be factored in the above manner.
For each of the following problems,
(a) Compute 2
4b ac .
(b) Use the information from part (a) to
determine whether or not the quadratic can
be written as factors with integer coefficients.
(Do not factor; simply answer Yes or No.)
1. 2 5 3x x
2. 2 7 10x x
3. 2 6 16x x
4. 2 6 4x x
5. 29 x
6. 27x x
7. 22 7 4x x
8. 26 1x x
9. 22 2 5x x
10. 25 4 1x x
Factor the following polynomials. If the polynomial
can not be rewritten as factors with integer
coefficients, then write the original polynomial as your
answer.
11. 2 4 5x x
12. 2 9 14x x
13. 2 5 6x x
14. 2 6x x
15. 2 7 12x x
16. 2 8 15x x
17. 2 12 20x x
18. 2 7 18x x
19. 2 5 24x x
20. 2 9 36x x
21. 2 16 64x x
22. 2 6 9x x
23. 2 15 56x x
24. 2 6 27x x
25. 2 11 60x x
26. 2 19 48x x
27. 2 17 42x x
28. 2 12 64x x
29. 2 49x
30. 2 36x
31. 2 3x
32. 2 8x
33. 29 25x
34. 216 81x
35. 22 5 3x x
36. 23 16 15x x
37. 28 2 3x x
38. 24 16 15x x
39. 29 9 4x x
40. 25 17 6x x
41. 24 3 10x x
42. 29 21 10x x
43. 212 17 6x x
44. 28 26 7x x
Factor the following. Remember to first factor out the
Greatest Common Factor (GCF) of the terms of the
polynomial, and to factor out a negative if the leading
coefficient is negative.
45. 2 9x x
46. 2 16x x
47. 25 20x x
Exercise Set A.1: Factoring Polynomials
University of Houston Department of Mathematics 708
48. 24 28x x
49. 22 18x
50. 28 8x
51. 4 25 20x x
52. 33 75x x
53. 22 10 8x x
54. 23 12 63x x
55. 210 10 420x x
56. 24 40 100x x
57. 3 29 22x x x
58. 3 27 6x x x
59. 3 24 4x x x
60. 5 4 310 21x x x
61. 4 3 26 6x x x
62. 3 22 80x x x
63. 5 39 100x x
64. 12 1049 64x x
65. 250 55 15x x
66. 230 24 72x x
Factor the following polynomials. (Hint: Factor first
by grouping, and then continue to factor if possible.)
67. 3 22 25 50x x x
68. 3 23 4 12x x x
69. 3 25 4 20x x x
70. 3 29 18 25 50x x x
71. 3 24 36 9x x x
72. 3 29 27 4 12x x x
APPENDIX A.2 Dividing Polynomials
MATH 1330 Precalculus 709
Appendix A.2: Dividing Polynomials
Polynomial Long Division and Synthetic Division
Polynomial Long Division and Synthetic Division
Long Division of Polynomials:
Example:
Solution:
APPENDICES
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APPENDIX A.2 Dividing Polynomials
MATH 1330 Precalculus 711
APPENDICES
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APPENDIX A.2 Dividing Polynomials
MATH 1330 Precalculus 713
APPENDICES
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Synthetic Division of Polynomials:
Example:
Solution:
APPENDIX A.2 Dividing Polynomials
MATH 1330 Precalculus 715
A Comparison of Long Division and Synthetic Division
Let us now analyze the previous two examples, both of which solved the same problem using
long division and then synthetic division.
Long Division Synthetic Division
4 3 22 05 8 3 5x x x x x
Constant: 5
Change the sign of the constant term when
performing synthetic division.
4 3 22 05 8 3 5x x x x x
Notice the coefficients of the dividend: 2, 0, 8, 3, 5
Write the coefficients of the dividend (without
changing any signs). Do not forget the
‘placeholder’ for 30x .
Notice that the coefficients in each column of
the subtraction problems under the division
sign (at the left) are similar to the numbers in
each column of the synthetic division problem
(above). Remember that at the left, the signs
are changed when the expressions are
subtracted.
3 2
4 3 2
4 3
3 2
3 2
2
2
2 10 42 213
5 2 0 8 3 5
2
10 8
10
4
10
2 3
42
213 5
213
1
50
210
106
060
5
x x x
x x x x x
x x
x x
x x
x x
x x
x
x
10 50 210 1
5 | 2 0 8 3 5
2 10 42 213 1060
065
2 0 8| 55 3
| 2 0 3 55 8
APPENDICES
University of Houston Department of Mathematics 716
Notice that the numbers in the answer line of
the synthetic division problem are the same as
the coefficients of the quotient plus the final
remainder in the long division problem.
In the long division problem, there is one
column for each power of x, and the
arithmetic in each column is done with the
coefficients.
Synthetic division is a shortcut for doing the
arithmetic with the coefficients without having
to write down all the variables. Remember that
this synthetic division procedure ONLY works
when the divisor is of the form D x x c .
The Remainder Theorem:
3 2
4 3 2
4 3
3 2
3 2
2
2
5 2 0 8 3 5
2 10
10 8
10 50
42 3
42
2 10 42 213
1060
210
213 5
213 1065
x x x
x x x x x
x x
x x
x x
x x
x x
x
x
5 | 2 0 8 3 5
10 50 210 1
2 10 42 213 1 6
065
0 0
APPENDIX A.2 Dividing Polynomials
MATH 1330 Precalculus 717
Additional Example 1:
Solution:
Additional Example 2:
Solution:
APPENDICES
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Additional Example 3:
Solution:
APPENDIX A.2 Dividing Polynomials
MATH 1330 Precalculus 719
Additional Example 4:
Solution:
APPENDICES
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APPENDIX A.2 Dividing Polynomials
MATH 1330 Precalculus 721
Additional Example 5:
Solution:
Exercise Set A.2: Dividing Polynomials
University of Houston Department of Mathematics 722
Use long division to find the quotient and the
remainder.
1. 2 6 11
2
x x
x
2. 2 5 12
3
x x
x
3. 2 7 2
1
x x
x
4. 2 6 5
4
x x
x
5. 3 22 19 12
3
x x x
x
6. 5
33222 23
x
xxx
7. 12
12656 23
x
xxx
8. 3 212 13 22 14
3 4
x x x
x
9. 3 2
2
2 13 28 21
3 1
x x x
x x
10. 4 3 2
2
7 4 42 12
7 2
x x x x
x x
11. 64
144433222
2345
x
xxxx
12. 362
4282201024
23468
xx
xxxxx
13. 5
1532
34
x
xx
14. xx
xxx
2
72432
35
Use synthetic division to find the quotient and the
remainder.
15. 2 8 4
10
x x
x
16. 3
642
x
xx
17. 5
286133 23
x
xxx
18. 4
312 23
x
xxx
19. 1
43 24
x
xx
20. 1
8732 45
x
xxx
21. 5
101827113 234
x
xxxx
22. 2
1251832 234
x
xxxx
23. 2
83
x
x
24. 3
814
x
x
25. 21
3 574
x
xx
26. 31
234 9106
x
xxx
Evaluate P(c) using the following two methods:
(a) Substitute c into the function.
(b) Use synthetic division along with the
Remainder Theorem.
27. 2;254)( 23 cxxxxP
28. 1;3875)( 23 cxxxxP
Exercise Set A.2: Dividing Polynomials
MATH 1330 Precalculus 723
29. 1;12487)( 23 cxxxxP
30. 3;14672)( 34 cxxxxP
Evaluate P(c) using synthetic division along with the
Remainder Theorem. (Notice that substitution without
a calculator would be quite tedious in these examples,
so synthetic division is particularly useful.)
31. 5;321703883)( 23567 cxxxxxxP
32. 2;11235103)( 2456 cxxxxxxP
33. 43234 ;12254)( cxxxxP
34. 273456 ;135932196)( cxxxxxxP
When the remainder is zero, the dividend can be
written as a product of two factors (the divisor and the
quotient), as shown below.
30
65 , so 30 5 6 .
2
62
3
x xx
x
, so 2
6 3 2x x x x
In the following examples, use either long division or
synthetic division to find the quotient, and then write
the dividend as a product of two factors.
35. 2 11 24
8
x x
x
36. 2 3 40
5
x x
x
37. 2 7 18
2
x x
x
38. 2 10 21
3
x x
x
39. 24 25 21
7
x x
x
40. 23 22 24
6
x x
x
41. 22 7 5
1
x x
x
42. 25 4 12
2
x x
x
APPENDICES
University of Houston Department of Mathematics 724
Appendix A.3: Geometric Formulas
Geometric Formulas
Geometric Formulas
The following two pages contain geometric formulas which may be helpful to you in this course.
Appendix A.3: Geometric Formulas
MATH 1330 Precalculus 725
s
h
r
h
w
r
h
b
h
b
1b
h
2b
r
d
s
w
Rectangle
Perimeter: 2 2P w
Area: A w
______________________________________________
Square
Perimeter: 4P s
Area: 2A s
______________________________________________
Parallelogram
Perimeter: Add the side lengths
Area: A bh
______________________________________________
Triangle
Perimeter: Add the side lengths
Area: 2
bhA
______________________________________________
Equilateral Triangle
Perimeter: 3P s
Area: 2 3
4
sA
______________________________________________
Circle
Circumference: 2C r d
Area: 2A r
______________________________________________
Trapezoid
Perimeter: Add the side lengths
Area: 1 2
2
b b hA
______________________________________________
Right Circular Cylinder
Lateral Area: 2L rh Ch
Total Surface Area: 2S L B ,
where B represents the area of
the base , so 22 2S rh r
Volume: 2V Bh r h
______________________________________________
Right Circular Cone
Lateral Area: 2
CL r
Total Surface Area: S L B ,
where B represents the area of
the base, so 2S r r
Volume: 2
3 3
Bh r hV
______________________________________________
Rectangular Prism
Lateral Area: 2 2L h wh Ph ,
where P represents the
perimeter of the base.
Total Surface Area: 2S L B ,
where B represents the area of
the base, so
2 2 2S h wh w
Volume: V Bh wh
______________________________________________
Sphere
Surface Area: 24S r
Volume: 34
3
rV
r
h
Appendix A.3: Geometric Formulas
University of Houston Department of Mathematics 726
c a
b
45o
45o
x
x
2x
x
1 1,x y
2 2,x y y
,M x y
60o
30o
x
3x 2x
d
x 1 1,x y
2 2,x y y
Pythagorean Theorem
2 2 2a b c
______________________________________________
Distance Formula
Distance between the points 1, 1x y and 2 2,x y :
2 2
2 1 2 1d x x y y
______________________________________________
Midpoint Formula
Midpoint of the segment joining the points 1, 1x y and
2 2,x y :
1 2 1 2, ,2 2
x x y yM x y
______________________________________________
30o-60
o-90
o Triangle
In a 30o-60o-90o triangle, the length of
the hypotenuse is twice the length of
the shorter leg, and the length of the
longer leg is 3 times the length of
the shorter leg.
______________________________________________
45o-45
o-90
o Triangle
In a 45o-45o-90o triangle, the legs
are congruent, and the length of the
hypotenuse is 2 times the length
of either leg.
______________________________________________
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