Question
Factor completely 2x2 − 18.
Prime
2(x2 − 9)
2(x + 3)(x − 3)
2(x + 3)(x + 3)
Question 2(Multiple Choice Worth 1 points)
(07.04 LC)
Factor completely 4x2 − 81.
(4x − 9)(x + 9)
(2x + 9)(2x + 9)
(2x + 9)(2x − 9)
(2x − 9)(2x − 9)
Question 3(Multiple Choice Worth 1 points)
(07.04 LC)
Factor completely x2 + 25.
(x + 5)(x + 5)
(x + 5)(x − 5)
(x − 5)(x − 5)
Prime
Question 4(Multiple Choice Worth 1 points)
(07.04 MC)
Factor completely 3x4 − 48.
3(x2 − 4)(x2 + 4)
3(x − 2)(x + 2)(x + 2)(x + 2)
3(x − 2)(x + 2)(x2 + 4)
3(x − 2)(x + 2)(x2 − 4)
Question 5(Multiple Choice Worth 1 points)
(07.04 MC)
Which shows 232 − 172 being evaluated using the difference of perfect squares method?
232 − 172 = (529 + 289)(529 − 289) = 196,320
232 − 172 = 529 − 289 = 240
232 − 172 = (23 + 17)(23 − 17) = (40)(6) = 240
232 − 172 = (23 − 17)2 = (6)2 = 36
Question 6(Multiple Choice Worth 1 points)
(07.04 LC)
Factor completely x2 − 49.
(x − 7)(x − 7)
(x + 7)(x − 7)
(x + 7)(x + 7)
Prime
Question 7(Multiple Choice Worth 1 points)
(07.04 MC)
Given the polynomial 6x3 + 4x2 − 6x − 4, what is the value of the constant 'k' in the factored form?
6x3 + 4x2 − 6x − 4 = 2(x + k)(x − k)(3x + 2)
0
1
2
4
Question 8(Multiple Choice Worth 1 points)
(07.04 MC)
Factor completely 625x8 − 1.
(25x4 − 1)(25x4 + 1)
(5x2 − 1)(5x2 + 1)(25x4 − 1)
(5x2 − 1)(5x2 + 1)(5x2 + 1)(5x2 + 1)
(5x2 − 1)(5x2 + 1)(25x4 + 1)
Question 9(Multiple Choice Worth 1 points)
(07.04 MC)
Factor completely 9x3 + 36x2 − x − 4.
(3x + 4)(3x − 4)(x + 1)
(3x + 1)(3x − 1)(x + 4)
(9x2 − 1)(x + 4)
(3x + 1)(3x − 1)(x − 4)
Question 10(Multiple Choice Worth 1 points)
(07.04 MC)
Factor completely 3x2 − 21.
3(x2 − 7)
3(x + 7)(x − 7)
3(x + 7)(x − 3)
Prime
Prime
2(x2 − 9)
2(x + 3)(x − 3)
2(x + 3)(x + 3)
Question 2(Multiple Choice Worth 1 points)
(07.04 LC)
Factor completely 4x2 − 81.
(4x − 9)(x + 9)
(2x + 9)(2x + 9)
(2x + 9)(2x − 9)
(2x − 9)(2x − 9)
Question 3(Multiple Choice Worth 1 points)
(07.04 LC)
Factor completely x2 + 25.
(x + 5)(x + 5)
(x + 5)(x − 5)
(x − 5)(x − 5)
Prime
Question 4(Multiple Choice Worth 1 points)
(07.04 MC)
Factor completely 3x4 − 48.
3(x2 − 4)(x2 + 4)
3(x − 2)(x + 2)(x + 2)(x + 2)
3(x − 2)(x + 2)(x2 + 4)
3(x − 2)(x + 2)(x2 − 4)
Question 5(Multiple Choice Worth 1 points)
(07.04 MC)
Which shows 232 − 172 being evaluated using the difference of perfect squares method?
232 − 172 = (529 + 289)(529 − 289) = 196,320
232 − 172 = 529 − 289 = 240
232 − 172 = (23 + 17)(23 − 17) = (40)(6) = 240
232 − 172 = (23 − 17)2 = (6)2 = 36
Question 6(Multiple Choice Worth 1 points)
(07.04 LC)
Factor completely x2 − 49.
(x − 7)(x − 7)
(x + 7)(x − 7)
(x + 7)(x + 7)
Prime
Question 7(Multiple Choice Worth 1 points)
(07.04 MC)
Given the polynomial 6x3 + 4x2 − 6x − 4, what is the value of the constant 'k' in the factored form?
6x3 + 4x2 − 6x − 4 = 2(x + k)(x − k)(3x + 2)
0
1
2
4
Question 8(Multiple Choice Worth 1 points)
(07.04 MC)
Factor completely 625x8 − 1.
(25x4 − 1)(25x4 + 1)
(5x2 − 1)(5x2 + 1)(25x4 − 1)
(5x2 − 1)(5x2 + 1)(5x2 + 1)(5x2 + 1)
(5x2 − 1)(5x2 + 1)(25x4 + 1)
Question 9(Multiple Choice Worth 1 points)
(07.04 MC)
Factor completely 9x3 + 36x2 − x − 4.
(3x + 4)(3x − 4)(x + 1)
(3x + 1)(3x − 1)(x + 4)
(9x2 − 1)(x + 4)
(3x + 1)(3x − 1)(x − 4)
Question 10(Multiple Choice Worth 1 points)
(07.04 MC)
Factor completely 3x2 − 21.
3(x2 − 7)
3(x + 7)(x − 7)
3(x + 7)(x − 3)
Prime
Answers
Answered by
GPT-4o mini
Here are the answers to your questions with explanations for the factorization:
**Question 1: Factor completely \( 2x^2 - 18 \).**
- Factor out the common factor 2:
\[ 2(x^2 - 9) \]
- Further factor \( x^2 - 9 \) as a difference of squares:
\[ 2(x + 3)(x - 3) \]
- **Answer: 2(x + 3)(x - 3)**
---
**Question 2: Factor completely \( 4x^2 - 81 \).**
- Recognize this as a difference of squares:
\[ (2x)^2 - 9^2 \]
- Factor it:
\[ (2x - 9)(2x + 9) \]
- **Answer: (2x - 9)(2x + 9)**
---
**Question 3: Factor completely \( x^2 + 25 \).**
- This is not factorable over the reals (it is a sum of squares).
- **Answer: Prime**
---
**Question 4: Factor completely \( 3x^4 - 48 \).**
- Factor out the common factor 3:
\[ 3(x^4 - 16) \]
- Recognize \( x^4 - 16 \) as a difference of squares:
\[ (x^2 - 4)(x^2 + 4) \]
- Further factor \( x^2 - 4 \) as a difference of squares:
\[ (x - 2)(x + 2) \]
- Thus, complete factorization is:
\[ 3(x - 2)(x + 2)(x^2 + 4) \]
- **Answer: 3(x - 2)(x + 2)(x^2 + 4)**
---
**Question 5: Which shows \( 23^2 - 17^2 \) being evaluated using the difference of perfect squares method?**
- Use the difference of squares:
\[ a^2 - b^2 = (a - b)(a + b) \]
- Thus:
\[ 23^2 - 17^2 = (23 - 17)(23 + 17) = (6)(40) = 240 \]
- **Answer: 232 − 172 = (23 + 17)(23 − 17) = (40)(6) = 240**
---
**Question 6: Factor completely \( x^2 - 49 \).**
- Recognize this as a difference of squares:
\[ x^2 - 7^2 = (x - 7)(x + 7) \]
- **Answer: (x + 7)(x - 7)**
---
**Question 7: Given the polynomial \( 6x^3 + 4x^2 - 6x - 4 \), find the value of 'k'.**
- Factor by grouping:
\[ 2x^2(3x + 2) - 2(3x + 2) = (3x + 2)(2x^2 - 2) \]
- Factor out the common factor:
\[ (3x + 2)2(x^2 - 1) = (3x + 2)2(x - 1)(x + 1) \]
- Thus, comparing with \( 2(x + k)(x - k)(3x + 2) \), \( k \) must be 1.
- **Answer: 1**
---
**Question 8: Factor completely \( 625x^8 - 1 \).**
- Recognize this as a difference of squares:
\[ (25x^4)^2 - 1^2 = (25x^4 - 1)(25x^4 + 1) \]
- Further factor \( 25x^4 - 1 \):
\[ 25x^4 - 1 = (5x^2 - 1)(5x^2 + 1)(5x^2 + 1) \]
- **Answer: (5x^2 - 1)(5x^2 + 1)(5x^2 + 1)(5x^2 + 1)**
---
**Question 9: Factor completely \( 9x^3 + 36x^2 - x - 4 \).**
- Factor by grouping:
\[ (9x^3 + 36x^2) + (-x - 4) = 9x^2(x + 4) - 1(x + 4) \]
- This gives:
\[ (x + 4)(9x^2 - 1) \]
- Also recognize \( 9x^2 - 1 \) as a difference of squares:
\[ (3x - 1)(3x + 1) \]
- Thus:
\[ (x + 4)(3x - 1)(3x + 1) \]
- **Answer: (3x + 1)(3x - 1)(x + 4)**
---
**Question 10: Factor completely \( 3x^2 - 21 \).**
- Factor out the common factor:
\[ 3(x^2 - 7) \]
- Thus,
- **Answer: 3(x^2 - 7)**
Feel free to reach out if you have more questions!
**Question 1: Factor completely \( 2x^2 - 18 \).**
- Factor out the common factor 2:
\[ 2(x^2 - 9) \]
- Further factor \( x^2 - 9 \) as a difference of squares:
\[ 2(x + 3)(x - 3) \]
- **Answer: 2(x + 3)(x - 3)**
---
**Question 2: Factor completely \( 4x^2 - 81 \).**
- Recognize this as a difference of squares:
\[ (2x)^2 - 9^2 \]
- Factor it:
\[ (2x - 9)(2x + 9) \]
- **Answer: (2x - 9)(2x + 9)**
---
**Question 3: Factor completely \( x^2 + 25 \).**
- This is not factorable over the reals (it is a sum of squares).
- **Answer: Prime**
---
**Question 4: Factor completely \( 3x^4 - 48 \).**
- Factor out the common factor 3:
\[ 3(x^4 - 16) \]
- Recognize \( x^4 - 16 \) as a difference of squares:
\[ (x^2 - 4)(x^2 + 4) \]
- Further factor \( x^2 - 4 \) as a difference of squares:
\[ (x - 2)(x + 2) \]
- Thus, complete factorization is:
\[ 3(x - 2)(x + 2)(x^2 + 4) \]
- **Answer: 3(x - 2)(x + 2)(x^2 + 4)**
---
**Question 5: Which shows \( 23^2 - 17^2 \) being evaluated using the difference of perfect squares method?**
- Use the difference of squares:
\[ a^2 - b^2 = (a - b)(a + b) \]
- Thus:
\[ 23^2 - 17^2 = (23 - 17)(23 + 17) = (6)(40) = 240 \]
- **Answer: 232 − 172 = (23 + 17)(23 − 17) = (40)(6) = 240**
---
**Question 6: Factor completely \( x^2 - 49 \).**
- Recognize this as a difference of squares:
\[ x^2 - 7^2 = (x - 7)(x + 7) \]
- **Answer: (x + 7)(x - 7)**
---
**Question 7: Given the polynomial \( 6x^3 + 4x^2 - 6x - 4 \), find the value of 'k'.**
- Factor by grouping:
\[ 2x^2(3x + 2) - 2(3x + 2) = (3x + 2)(2x^2 - 2) \]
- Factor out the common factor:
\[ (3x + 2)2(x^2 - 1) = (3x + 2)2(x - 1)(x + 1) \]
- Thus, comparing with \( 2(x + k)(x - k)(3x + 2) \), \( k \) must be 1.
- **Answer: 1**
---
**Question 8: Factor completely \( 625x^8 - 1 \).**
- Recognize this as a difference of squares:
\[ (25x^4)^2 - 1^2 = (25x^4 - 1)(25x^4 + 1) \]
- Further factor \( 25x^4 - 1 \):
\[ 25x^4 - 1 = (5x^2 - 1)(5x^2 + 1)(5x^2 + 1) \]
- **Answer: (5x^2 - 1)(5x^2 + 1)(5x^2 + 1)(5x^2 + 1)**
---
**Question 9: Factor completely \( 9x^3 + 36x^2 - x - 4 \).**
- Factor by grouping:
\[ (9x^3 + 36x^2) + (-x - 4) = 9x^2(x + 4) - 1(x + 4) \]
- This gives:
\[ (x + 4)(9x^2 - 1) \]
- Also recognize \( 9x^2 - 1 \) as a difference of squares:
\[ (3x - 1)(3x + 1) \]
- Thus:
\[ (x + 4)(3x - 1)(3x + 1) \]
- **Answer: (3x + 1)(3x - 1)(x + 4)**
---
**Question 10: Factor completely \( 3x^2 - 21 \).**
- Factor out the common factor:
\[ 3(x^2 - 7) \]
- Thus,
- **Answer: 3(x^2 - 7)**
Feel free to reach out if you have more questions!
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