1 – 2x + x2 x2 – 1 2c2 + 5c – 25 14c – 21, algebra homework help

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Mathematics

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In this discussion, you are assigned two rational expressions to work on. Remember to factor all polynomials completely. Read the following instructions in order and view the example to complete this discussion. Please complete the following problems according to your assigned number. (Instructors will assign each student their number.)

If your assigned number is:

Your first rational expression is

Your second rational expression is

1

a2 + 2a – 10
–11a

2x – 8
x2 – x

2

y2 + 11y + 30
y + 5

5a – 3
a2 – 49

3

u2 + u – 72
3u – 15

2w – 1
w2 – 9

4

x2 – 6x – 55
x2 – 3x – 28

b – 3
4b2 – 16

5

16b2 – 1
2b2 + 11b – 6

c2 + 2c – 10
10c2 – 80c + 160

6

4w2 – 9
–5w

y2 – 4
2y2 + y – 1

7

a2 + 6a + 9
a – 10

t2 + 2t – 10
t2 – 6t + 8

8

81m2 – 16
5m + 10

22x + 11
2x2 – x

9

3a2 + 16a + 5
a2 – 7a + 10

b2 + 7b + 10
b2 – 9

10

4x2 – 1
6x – 18

r2 + 2r – 24
r – 2

11

t2 – 14t + 49
4t – 8

3q2 – 22q + 24
q2 – 1

12

v2 – v – 20
v2 + 4v + 3

6t – 3
t

13

n2 – 2n – 15
n – 2

8k + 6
8k2 + 2k – 3

14

64k2 – 9
2

s2 + 2s – 15
s2 – 36

15

4k2 – 12k + 9
4k

3w2 + 36
2w – 8

16

5m2 + m
m + 6

12a – 15
5a – 25

17

7t2 – 14t
4t2 – 9

36 – w2
w

18

22x + 11
x2 – 3x – 10

1 – 2c
20c2 + 10c

19

20c3 + 5c2 – c
7c – 14

3a2 – 12a
24a2 – 18a

20

g2 – 36
6g2 + 15g

7n – 2
4n2 – 25

21

1 – 2x + x2
x2 – 1

2c2 + 5c – 25
14c – 21

22

144 – w2
8 – 2w

z2 – 8z + 16
–12

23

4 + 16n2
n – 8

3x2 – 2x – 1
x2 – 81

24

y2 – 25
–6y

37
2p – 4p2

25

a2 – 100
a2 + 5a + 6

3b2 – 9
b – 8

26

1 – x2
x2 + 10x + 25

2u – 2
1 – u

27

3z + 3
3z

5y3 – 75y
2y2 + y – 15

28

9 – 36x2
8

15k2 – 5k
k2 – k – 30

29

x2 – 7x + 12
5x

w2 – 9w – 36
16w2 – 1

30

m2 + 13m + 40
6m

4y – 3
25y2 – 4

31

k2 – 8k
17

2b + 1
3b2 – 12

32

9b2 + 3
41

2x – 6
10x2 + 5x

33

15x2 + 45
50x

42
m2 – 3m

34

g2 + 46g
g

3k + 1
k2 + k – 42

35

4a – 5
14

4m3 +16m
3m2 – m

36

x2 – 25
23

b2 – 18b + 81
3b2 – 12

37

9m2 – 4
23

5x + 15
x2 – 49

38

d2 + 9
33

m2 + 4m – 5
5m2 + m

39

13n2 – 13n
6n

2w + 1
9w2 – 1

40

4x3 – 16
x

14
2b2 – 8

41

x2 + 2x – 10
3x

2x – 8
x2 – 7x + 10

42

x2 + x – 72
24

5b – 3
b2 – 4

43

w2 + 11w + 30
7w

2n – 1
4n2 – 9

44

g2 – 6g – 55
g

k3 + k
k2 – k – 42

45

16t2 – 1
64

x2 + 2x – 10
3x – 15

  • Explain in your own words what the meaning of domain is. Also, explain why a denominator cannot be zero.
  • Find the domain for each of your two rational expressions.
  • Write the domain of each rational expression in set notation (as demonstrated in the example).
  • Do both of your rational expressions have excluded values in their domains? If yes, explain why they are to be excluded from the domains. If no, explain why no exclusions are necessary.
  • Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing. Do not write definitions for the words; use them appropriately in sentences describing your math work.
    • Domain
    • Excluded value
    • Set
    • Factor
    • Real numbers

Your initial post should be at least 250 words in length. Support your claims with examples from required material(s) and/or other scholarly resources, and properly cite any references.

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Explanation & Answer

Here is my answer :)

Algebra Question
Assigned number is 21
1 − 2𝑥 + 𝑥 2
2c 2 + 5c − 25
𝑎𝑛𝑑
𝑥2 − 1
14c − 21

First rational expression:
1 − 2𝑥 + 𝑥 2
𝑥2 − 1
𝑥 2 − 2𝑥 + 1
𝑥2 − 1
Factor all polynomials
𝑥 2 − 2𝑥 + 1
𝑎 = 1, 𝑏 = −2, 𝑐 = 1
𝑥=

𝑥=

−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎

−(−2) ± √(−2)2 − 4(1)(1)
2(1)
𝑥=

2 ± √4 − 4
2

𝑥=

2 ± √0
2

𝑥=

2
=1
2

So,
𝑥 2 − 2𝑥 + 1 = (𝑥 − 1)2

Now,
𝑥 2 − 1 = (𝑥 − 1)(𝑥 + 1)
So,

𝑥 2 − 2𝑥 + 1
(𝑥 − 1)2
=
(𝑥 − 1)(𝑥 + 1)
𝑥2 − 1
𝑥 2 − 2𝑥 + 1 (𝑥 − 1)
=
𝑥2 − 1
(𝑥 + 1)
Second rational expression:
2c 2 + 5c − 25
14c − 21
Factor all polynomials

2c 2 + 5c − 25
𝑎1 = 2, 𝑏1 = 5, 𝑐1 = −25
−𝑏1 ± √𝑏1 2 − 4𝑎1 𝑐1
𝑐=

𝑐=

2𝑎1

−5 ± √(5)2 − 4(2)(−25)
2(2)
𝑐=

−5 ± √25 + 200
4

𝑐=

−5 ± √225
4

𝑐=
𝑐𝐹1 =

−5 ± 15
4

−5 + 15 10 5
=
=
4
4
2
𝑐𝐹1 =

𝑐𝐹2 =

5
2

−5 − 15 −20
=
= −5
4
4
𝑐𝐹2 = −5

So,

5
2c 2 + 5c − 25 = (c + 5) (c − )
2
Now,

14c − 21 = 14 (c −

21
)
14

3
14c − 21 = 14 (c − )
2
So,
5
2c 2 + 5c − 25 (c + 5) (c − 2)
=
3
14c − 21
14 (c − 2)







Explain in your own words what the meaning of domain is. Also, explain why
a denominator cannot be zero.
Find the domain for each of your two rational expressions.
Write the domain of each rational expression in set notation (as
demonstrated in the example).
Do both of your rational expressions have excluded values in their domains?
If yes, explain why they are to be exclud...


Anonymous
I was having a hard time with this subject, and this was a great help.

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