Algebra is a branch of mathematics that deals with the study of rules of operations and relations. Diophantus is regarded as the father of algebra. Basic algebraic concepts include variables and constants, expressions, terms, polynomials, equations and algebraic structures. So much is known about the subject that study of this takes a life time. Algebra offers lot of food for thought for students and mathematics lovers will have lot of fun solving problems in algebra. The beauty of the subject is that it prompts the student to think indicatively to solve problems and helps to develop analytical and logical skills for the student. Study of mathematics at any level will cover some topics in algebra.
High school algebra covers topics such as polynomials, algebraic expressions and identities, equtaions and factorization.
Let us learn some solved problems in the above topics of high school algebra.
Learning Polynomials and Factorization in high school algebra:
At high school level, polynomials and factorization are important topics covered in algebra.
POLYNOMIALS:
An Algebraic expression of the form axn is called Monomial in x. For example, 7x3 .The sum of two monomials are called a Binomial and the sum of three monomials are called Trinomial. For example, 2x3 + 3x is a binomial and 2x5 – 3x2 + 3 is Trinomial. The sum of a finite number of monomials in x is called a polynomial in x.
Example:
Find the sum of 2x4 – 3x2 + 5x + 3 and 4x + 6x3 – 6x2 – 1.
Solution:
Using the associative and distributive properties of 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. (answer)
FACTORIZATION :
The process of writing polynomial as a product of two or more simpler polynomials is called Factorization.
The way of writing a polynomial as a product of two or more simpler polynomials is called factorization. The process of factorization is also known as the resolution into factors.
Example 1: Factorize x2 – 2xy – x + 2y.
Solution:
x2 – 2xy – x + 2y = (x2 – 2xy) – (x – 2y)
= x(x – 2y) + (–1) (x – 2y)
= (x – 2y) [x + (–1)]
= (x – 2y) (x – 1). (answer)
Example 2: Solve 9x2 = x.
Solution:
9x2 = x
9x2 − x = 0
x (9x − 1) = 0 ab = 0
x = 0 or x =
What if we attempt the same problem using the another method?
9x2 = x
9x = 1 divide by x
x =
Every quadratic equations has 2 roots. Dividing the quadratic equation by x removes the root x = 0.
However, dividing by a constant does not impact the roots.
Learning Algebraic Identites in high school algebra:
Algebraic Identities:
Algebraic identities is important in High school algebra . Algebraic identities are algebraic equations satisfied by any value of the variables.
The algebraic identity of x + 0 = x tells us that anything (x) added to zero equals the original "anything," no matter what rate that "anything" (x) may be. Like normal algebra, Boolean algebra has its own unique identities base on the bivalent states of Boolean variables.
Example 4: Solve for variables x,y and z:
x + y + 2z = 2 ------> (1)
3x - y + 3z = 4 ------>(2)
2x + y + 4z = 6 ------>(3)
Solution:
Solve (1) and (2),
x + y + 2z = 2 ----> (1)
3x - y + 3z = 4 -----> (2)
add the above two equations.
we get 4x + 5z = 6 ------> (4)
solve (1) and (3)
(3) * 2 ----> 4x + 4y +8z = 12
(4) -----> 4x + 0y +5z = 6
(-) (-) (-)
3z = 6
z = 2
substitute z = 2 in (4) eqn
4x + 5(2) = 6
4x = -4
x = -1
substitute x = -1, z = 2 in (1) eqn.
(-1) + y + 2(2) = 2
y -3 = 2
y = 5.
x = -1, y = 5, z = 2.
High school algebra covers topics such as polynomials, algebraic expressions and identities, equtaions and factorization.
Let us learn some solved problems in the above topics of high school algebra.
Learning Polynomials and Factorization in high school algebra:
At high school level, polynomials and factorization are important topics covered in algebra.
POLYNOMIALS:
An Algebraic expression of the form axn is called Monomial in x. For example, 7x3 .The sum of two monomials are called a Binomial and the sum of three monomials are called Trinomial. For example, 2x3 + 3x is a binomial and 2x5 – 3x2 + 3 is Trinomial. The sum of a finite number of monomials in x is called a polynomial in x.
Example:
Find the sum of 2x4 – 3x2 + 5x + 3 and 4x + 6x3 – 6x2 – 1.
Solution:
Using the associative and distributive properties of 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. (answer)
FACTORIZATION :
The process of writing polynomial as a product of two or more simpler polynomials is called Factorization.
The way of writing a polynomial as a product of two or more simpler polynomials is called factorization. The process of factorization is also known as the resolution into factors.
Example 1: Factorize x2 – 2xy – x + 2y.
Solution:
x2 – 2xy – x + 2y = (x2 – 2xy) – (x – 2y)
= x(x – 2y) + (–1) (x – 2y)
= (x – 2y) [x + (–1)]
= (x – 2y) (x – 1). (answer)
Example 2: Solve 9x2 = x.
Solution:
9x2 = x
9x2 − x = 0
x (9x − 1) = 0 ab = 0
x = 0 or x =
What if we attempt the same problem using the another method?
9x2 = x
9x = 1 divide by x
x =
Every quadratic equations has 2 roots. Dividing the quadratic equation by x removes the root x = 0.
However, dividing by a constant does not impact the roots.
Learning Algebraic Identites in high school algebra:
Algebraic Identities:
Algebraic identities is important in High school algebra . Algebraic identities are algebraic equations satisfied by any value of the variables.
The algebraic identity of x + 0 = x tells us that anything (x) added to zero equals the original "anything," no matter what rate that "anything" (x) may be. Like normal algebra, Boolean algebra has its own unique identities base on the bivalent states of Boolean variables.
Example 4: Solve for variables x,y and z:
x + y + 2z = 2 ------> (1)
3x - y + 3z = 4 ------>(2)
2x + y + 4z = 6 ------>(3)
Solution:
Solve (1) and (2),
x + y + 2z = 2 ----> (1)
3x - y + 3z = 4 -----> (2)
add the above two equations.
we get 4x + 5z = 6 ------> (4)
solve (1) and (3)
(3) * 2 ----> 4x + 4y +8z = 12
(4) -----> 4x + 0y +5z = 6
(-) (-) (-)
3z = 6
z = 2
substitute z = 2 in (4) eqn
4x + 5(2) = 6
4x = -4
x = -1
substitute x = -1, z = 2 in (1) eqn.
(-1) + y + 2(2) = 2
y -3 = 2
y = 5.
x = -1, y = 5, z = 2.
No comments:
Post a Comment