## Algebra 2 a Semester Exam

**Introduction :**

Algebra is the branch of mathematics 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 number theory, algebra is one of the main branches of pure mathematics, in this topic we are discussed about algebra 2 a semester exam introduction and algebra 2 a semester exam example problems (source: Wikipedia).

Algebra 2 a Semester Exam Example: 1

**Consider the quadratic equation**

**x****2 − 4**

**x****+ 7 = 0**

**Solution:**

Its discriminate is

*b*2 − 4

*ac*= (− 4)2 − (4) (7) (1)

= 16 − 28 = − 12 ( -ve.)

∴ This equations are not real. The roots are given by

(− (− 4) ± `sqrt(-12)`)`/` 2 =(4 ± `sqrt(-12)` )`/` 2 = 2 ±

*i `sqrt(3)`*

Thus we see that the roots 2 +

*i*3 and 2 −

*i*3 are conjugate to each other.

Now consider it as cube roots of unity.

Let

*x*be the cube root of unity then

*x*= (1)1/3

⇒

*x*3 = 1

⇒ (

*x*− 1) (

*x*2 +

*x*+ 1) = 0

∴

*x*− 1 = 0 ;

*x*2 +

*x*+ 1 = 0

Hence

*x*= 1 and

*x*= (− 1 ± 1 −`sqrt(((4)(1)(11))/2)` )

∴ Cube roots of unity are 1,

(− 1 + `sqrt(3i)`)

*`/`*2 , (− 1 − `sqrt(3i)`

*)`/`*2

Here again the two complex roots

(− 1 +`sqrt(3i)`

*)*`/` 2 and (− 1 − `sqrt(3i)`

*)*`/` 2 are conjugate to each other.

Algebra 2 a Semester Exam Example: 2

**Multiply**

**(A + B)3**

**out.**

**Solution:**

Rewrite the value so you have something you multiply out.

(A + B)(A + B)(A + B)

Multiply the first two binomials together.

(A + B)(A + B)

A2 + AB + BA + B2

After combining like terms, you have

A2 + 2AB + B2

now You have a binomial and trinomial to multiply together.

(A2 + 2AB + B2)(A + B)

In this step slightly more complicated situations while multiplying a binomial by another binomial. So we are multiplying first term and trinomial then we are multiplying the last term with binomial of each term in the trinomials are shown below as like the equation.

A3 + 2A2B + AB2 + BA2 + 2AB2 + B3

Combine like terms if possible to simplify the answer.

A3 + 3A2B + 3AB2 + B3

Therefore the final multiplied value A3 + 3A2B + 3AB2 + B3