Abstract Algebra Homework Solutions

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Algebra is the branch of mathematics and that concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with geometry, analysis, topology, combinatorics, and also the number theory, algebra is one of the main branches of pure mathematics. Now we see about abstract algebra homework solutions to (Source : Wikipedia).

problems and Solutions in abstract algebra homework :

Example 1:

Abstract algebra homework solutions to Find the sum of 2x4 – 3x2 + 5x + 3 and 4x + 6x3 – 6x2 – 1.

Solution:

Using the associative and distributive properties of that real numbers, we obtain

(2x4 – 3x2 + 5x + 3) + (6x3 – 6x2 + 4x – 1) = 2x4 + 6x3 – 3x2 – 6x2 + 5x + 4x + 3 – 1

= 2x4 + 6x3 – (3+6)x2 + (5+4)x + 2

= 2x4 + 6x3 – 9x2 + 9x + 2.

The following scheme is helpful in adding two polynomials

2x4 + 0x3 – 3x2 + 5x + 3

0x4 + 6x3 – 6x2 + 4x – 1

____________________

2x4 + 6x3– 9x2 + 9x + 2

_____________________

Example ...
... 2:

Abstract algebra homework solutions to Subtract 2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6.

Solution:

Using the associative and distributive properties, we have

(x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1) = x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1

= x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1

= (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1)

= –x3 + 8x2 – 4x – 5

Example 3:

Abstract algebra homework solutions to Factorize 6x5y2 + 6x4y3 + 9x2y4 + 9xy5.

Solution :

Applying both step 1 and step 2, we have

6x5y2 + 6x4y3 + 9x2y4 + 9xy5 = 3xy2(2x4 + 2x3y + 3xy2 + 3y3)

= 3xy2 [(2x4 + 3xy2) + (2x3y + 3y3)] = 3xy2 [x(2x3 + 3y2) + y(2x3 + 3y2)]

= 3xy2 (2x3 + 3y2) (x + y).

Example 4:

Abstract algebra homework solutions to Factorize x2 – 2xy – x + 2y.

Solution:

The terms of the expression do not have a common factor. However, we observe that the terms can be grouped as follows:

x2 – 2xy – x + 2y = (x2 – 2xy) – (x – 2y) = x(x – 2y) + (–1) (x – 2y)

= (x – 2y) [x + (–1)] = (x – 2y) (x – 1).

Example 5: Solve

Practice problems in abstract algebra homework solutions

1. Factorize 6x4y3 – 4x2y2 + 10xy3.

Answer: = 2xy2(3x3y – 2x + 5y).

2. Solve the equations : x + 2y + 3z = 14, 3x + y + 2z = 11, 2x + 3y + z = 11

Answer: The solution is x = 1, y = 2, z = 3.

Abstract algebra is the subject area of arithmetic that studies algebraic structures, such as groups, rings, fields, modules, vector spaces, and algebras. The phrase conceptual algebra was coined at the turn of the 20th century to distinguish this area from what was normally referred to as algebra, the study of the rules for manipulating formulae and algebraic expressions involving unknowns and real or complex numbers, often now called elementary algebra. The distinction is rarely made in more recent writings.

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Abstract algebra problems explanations:

Here will explain about the Abstract algebra problems

Write a verbal expression for each algebraic expression.

1. Problems 1) using algebra equation

5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

Solution:-

Multiplying factors:

-15x - 10 - x + 3 = -16x - 20 +13

Group like terms:

-16x - 7 = -16x - 7

Add 16x+7 to both sides and using the equations as fallows

0 = 0

2. 4(xy) 2

solution:

Step1: first we take the root for find the factor

Step2: root on the both 4 and (xy) 2

Step3: finally we get the answer for 4(xy) 2

2xy *2xy=4(xy)2

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3.Simplify the algebraic expression 3(a -4) + 6b - 3(a -b -2) + 8

Solution:

Given the algebraic expression

3(a -4) + 6b - 3(a -b -2) + 8

Multiply factors.

= 3a - 12 + 6b -3a + 3b + 6 + 8

Grouping the above terms.

= 9b + 2

Write an algebraic expression for each verbal expression.

1. 6 more than twice m

Answer:

In this problem we write expression for the given condition

The value 6 is twice more than the variable m.

6>2m

2. 8 increased by three times a number

Answer:

In this problem we write expression for the given condition

The value 8 is improved with the three times of a variable m.

3n+8

Solving One-Step Equations

1.d - 8 = 17

Answer:

d=17+8 = d=25

2. V + 12 = -5

Answer:

v=-5-12 = -17

3. b - 2 = -11

Answer:

b=-11+2 = -9

4. -16 = m + 71

Answer:

-16-71=m;

m=-87

Abstract algebra problems for practice

Here will give some abstract algebra problems for learning

Example: 1

If x = 6, what is the value of x2 + 3x + 7?

Solution:

In this problem we study how to do a given expression x2 + 3x + 7, along with a value for x equal to 6. The exact method is to substitute the numerical value into the expression and then make sure to do the math correctly.

(6)2 + 3(6)+7 =

36 + 18 + 7 =61

Example: 2

4x - 5 = 2x – 2

Solution:

Add +5 on both sides

+5 = + 5

4x = 2x + 3

Add -2x on both sides

-2x= -2x

2x = 3

Divide by 2 on both sides

2x/2 =3/2

Answer X = 3/2

Example3:

15 less than k

Answer:

In this basic algebra problem we write expression for the given condition

The 15 is less than the variable k value

15 k

Example4:

(−13a +5b)+(18a +(−18b))

Solution:

= −13a +18a +5b + (−18b)
= 5a + (−13b)
= 5a −13b
by the commutative law, we may also write this as −13b +5a.

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