Exercise I: Given the following polynomials: A = 15x3y + 2xy3 – 5x²y? +4 B = -6x²y – 11xy + 4x?y? - 3 + xy Find: a) A+B b) A- B c) 2B + A

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Exercise I:
Given the following polynomials:
A = 15x?y + 2xy3 – 5x²y? +4
B = -6x²y – 11xy3 + 4x2y2 - 3 + xy
Find:
a) A+B
b) A – B
c) 2B + A
Exercise II:
Expand and reduce these expressions:
1) (3x – 2)(x – 1) + 2(x + 4)
2) (2x + 1)2 – (x + 2)(x – 2)
3) (x + 5)2 + (2x – 3)2
4) 2(x + 2)(x – 1) + (x – 3)(x + 3) – (x – 2)2
5) 4x(x – 6) + (2x – 1)(x – 3) +5
Exercise III:
Factorize these expressions:
1) 18x'y? - 36x²y³ + 9x?y – 27x?y
2) 25x? + 10x + 1
3) (2x – 3)2 – (x + 5)²
4) x2 – 4x + 4 - 3(x - 2)(x + 1)
5) (x + 5)(x – 4) – (4 - x)2 + (2x - 8)
6) x3 – 5x2 – x + 5
7) 4(x + 3)? – (x – 2)?
8) (3x + 6)(x – 1) + (x² + 4x + 4) – (x² – 4)
|
Exercise IV:
Solve the following equations:
1) 3x + 2(x + 3) – 2 = 2x + 5
2) (x + 3)(3x – 1) = (3x – 1)(2 - x)
3) (2x – 3)2 + 2(x + 1) = 4(x + 2)?
х-2
ド+さ+1
3
2
3x+1
- 1
2
5) (4x – 3)2 = (x + 1)?
6) (x + 4)(2x – 3)(x + 1) = 0
Transcribed Image Text:Exercise I: Given the following polynomials: A = 15x?y + 2xy3 – 5x²y? +4 B = -6x²y – 11xy3 + 4x2y2 - 3 + xy Find: a) A+B b) A – B c) 2B + A Exercise II: Expand and reduce these expressions: 1) (3x – 2)(x – 1) + 2(x + 4) 2) (2x + 1)2 – (x + 2)(x – 2) 3) (x + 5)2 + (2x – 3)2 4) 2(x + 2)(x – 1) + (x – 3)(x + 3) – (x – 2)2 5) 4x(x – 6) + (2x – 1)(x – 3) +5 Exercise III: Factorize these expressions: 1) 18x'y? - 36x²y³ + 9x?y – 27x?y 2) 25x? + 10x + 1 3) (2x – 3)2 – (x + 5)² 4) x2 – 4x + 4 - 3(x - 2)(x + 1) 5) (x + 5)(x – 4) – (4 - x)2 + (2x - 8) 6) x3 – 5x2 – x + 5 7) 4(x + 3)? – (x – 2)? 8) (3x + 6)(x – 1) + (x² + 4x + 4) – (x² – 4) | Exercise IV: Solve the following equations: 1) 3x + 2(x + 3) – 2 = 2x + 5 2) (x + 3)(3x – 1) = (3x – 1)(2 - x) 3) (2x – 3)2 + 2(x + 1) = 4(x + 2)? х-2 ド+さ+1 3 2 3x+1 - 1 2 5) (4x – 3)2 = (x + 1)? 6) (x + 4)(2x – 3)(x + 1) = 0
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