Math article on grade XI (Algebra ii – Operations on polynomials)
Friends! In previous article we had started the discussion of Algebra ii and Absolute value is the main scenario of previous article. If you get remember then in the end of previous article the discussion of System of equation was started and we will continue its remaining part in this article. Along with System of equation, today we will also explore various operations that are usually applied on polynomials.
In previous article we already had explored the Row reduction method of solving system of linear equations and now here we are going to discuss direct solving method for system of linear equation:
For solving this we will use two newer terms that are vector presentation of matrix and inverse matrix.
As explained in previous session, a system of linear equation can be represent as AX = C
Where A refers the coefficient matrix, x represents the variables derivatives and C represents the RHS of the system. But here we will use their vector form as;
AX(vector) = C(vector)
Here x is an n-dimensional vector the elements of which represent the solution of the equations and C (vector) is the constant vector of the system of equations. Now its solution can be written in the following form:
X = A-1 C
Here A-1, represents Inverse matrix of co-efficient of the system.
Suppose a system is as:
-x + 3y + z = 1
2x + 5y = 3
3x + y - 2z = -2
Now write this system into standard form of AX = C, as
-1 3 1 x 1
2 5 0 y 3
3 1 -2 z -2
Now find the inverse of the Matrix A, as
A-1 = -10/9 7/9 -5/9
4/9 -1/9 2/9
-13/9 10/9 -11/9
Now multiply both sides of equation by this inverse matrix:
On the left side, A-1 cancel out the matrix A, and on the right side:
= -10/9 7/9 -5/9 1
4/9 -1/9 2/9 3
-13/9 10/9 -11/9 -2
= 21
-3
39
This is the solution of the system of linear equations that can be represented as:
X 21
Y = -3
Z 39
This is the second way by which any system of linear equation can be sort out but one thing to notice that every defined method for solving this type of queries required the implementation of matrices and its application, so students has to make their command on all the fundamentals and operation procedures of matrices for solving queries like system of linear equations.
So Now let us begin with the topic: operations on Polynomials;
“What is a Polynomial and how it is formed?, this query is not required to be answer for grade XI students, because you guys should aware of this, before coming into this standard. So here we are directly starting the operations discussion of polynomials:
First arithmetic operation to be applied on polynomial is Addition: while using this operation, students need to implement two standard properties that are Additional property and Distributive property that implies to combine like terms, like terms refer to same variables and same exponents for the variables.
distributive property worksheets in Algebra: if an expression is as ba + ca then it can be rewritten as a(b + c)
ba + ca = a(b + c)
let us take an example to explain it more:
Suppose given polynomial is
(3X2 + 7x + 8) + (5x2 - 8x + 2) = 3x2 + 7x + 8 + 5x2 - 8x + 2
Now use the distributive property to combine like terms as
(3x2 + 5x2) + (7x - 8x) + (8 + 2)
8x2 – x + 10
Now move towards the next operation that is Subtraction:
This one is pretty similar as Addition operation but when two terms are added then the second polynomial terms are inversed or multiplied by minus (-) sign.
We are using the same example as above:
(3X2 + 7x + 8) + [- (5x2 - 8x + 2)]
Here the second polynomial is need to be subtracted from first one that’s why firstly the second polynomial is multiplied by negative sign and then the simple addition will be performed.
3X2 + 7x + 8 - 5x2 + 8x – 2
-2x2 + 15x + 6
Third general operation, which is used to solve polynomials, is multiplication. Let us find out the simplest way to execute this operation on polynomials. While evaluating these operation students need to use above principles of addition and subtraction along with distributive property and law of exponents. Let us take an example to see the practical implementation of multiplication in between two polynomials. For more information on polynomials visit here
Example: 3x(2x -5) = (3x)(2x) – 3x(5) = 6x2 – 15x
Let us take one more example where number of derivatives is more than previous one:
( 2x + 1) (3x - 4) = (2x)(3x) – (2x)4 + (1)3x - (1)4
= 6x2 - 8x + 3x – 4
= 6x2 – 5x – 4
The fourth operation to be used to solve higher order polynomials is power operation. There are various standard formulas that are need to be learnt before starting the evaluation of complex polynomials. Some of these standard formulas are as following:
(x + y)0 = 1
(x + y)1 = x + y
(x + y)2 = x2 + 2xy + b2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
Let us take an example to use some of the above formulas for solving polynomials:
(x2 + 2x – 3)2 = (x2 + 2x – 3) (x2 + 2x – 3)
= (x2) (x2) + (x2)(2x) - (x2)(3) + (x2)(2x) + (2x)(x2) - (2x)(3) – 3(x2) - (2x)(3) +9
= x4 + 2x3 – 3x2 + 4x2 – 3x2 + 2x3 – 6x - 3x2 – 6x + 9
= x4 + 4x3 – 2x2 - 12x + 9
This is how the polynomial multiplication sorted out. While solving this student should know that whenever the bases are same, that time exponents are added to form a single term; This basic rule is learned in earlier classes but still most of the times students get confused while tracing the solved answer of the query.
Now the final operation we are going to explore in today’s article for polynomials is Division:
This operation includes some conditions that are needed to be taking in count while dividing two polynomials, as follows:
As you guys know that polynomials have many types as Monomials, Binomials and trinomials, so this is the reason the scenario of division process varies.
1. First condition is: Monomial
Monomial
To sort out this law of exponents is required:
(24x4y2)/ (18x2y3) = 4x2/3y2
This one is simple because cancellation of terms gets easier because of Monomial.
2. Polynomial monomial:-
(4x5 – 8x3 + 12x2)/ 6x2
= 4x5/6x2 – 8x3/6x2 + 12x2/6x2
= 2x3/3 – 4x/3 + 2
3. Polynomial ÷ polynomial;- for this purpose students need to follow long division algorithm that includes following steps:
a. first write the dividend and divisor polynomials into their standard polynomial form after that use zero coefficients for powers of the variable that are missing in the dividend and divisor.
b. then Divide first term of the divisor into the first terms of the dividend. Put its quotient above term in the dividend.
c. in third step, Multiply quotient by all terms of the divisor and put the products under the appropriate terms of the dividend.
d. (change signs on bottom polynomial and add) Subtract and bring down remaining terms.
e. divide the remaining term by first term until the power of the divisor is larger than the power in the dividend
Let’s take an example to execute this algorithm to solve
divide x2 + 9x +14 by (x + 7)
This is all about arithmetic operations on polynomials. If there is any doubt remaining in your mind related to polynomials operations or system of linear equation then you guys have the option of using Online math tutoring service where Online expert math tutors are waiting to give you immediate assistance with detailed explanation by solving related queries. These online tutors are available for 24 hours in a day so students can interact with them whenever they want.
In upcoming posts we will discuss about Types of Events in Grade XI and isosceles triangle theorem. Visit our website for information on Karnataka board syllabus
Friends! In previous article we had started the discussion of Algebra ii and Absolute value is the main scenario of previous article. If you get remember then in the end of previous article the discussion of System of equation was started and we will continue its remaining part in this article. Along with System of equation, today we will also explore various operations that are usually applied on polynomials.
In previous article we already had explored the Row reduction method of solving system of linear equations and now here we are going to discuss direct solving method for system of linear equation:
For solving this we will use two newer terms that are vector presentation of matrix and inverse matrix.
As explained in previous session, a system of linear equation can be represent as AX = C
Where A refers the coefficient matrix, x represents the variables derivatives and C represents the RHS of the system. But here we will use their vector form as;
AX(vector) = C(vector)
Here x is an n-dimensional vector the elements of which represent the solution of the equations and C (vector) is the constant vector of the system of equations. Now its solution can be written in the following form:
X = A-1 C
Here A-1, represents Inverse matrix of co-efficient of the system.
Suppose a system is as:
-x + 3y + z = 1
2x + 5y = 3
3x + y - 2z = -2
Now write this system into standard form of AX = C, as
-1 3 1 x 1
2 5 0 y 3
3 1 -2 z -2
Now find the inverse of the Matrix A, as
A-1 = -10/9 7/9 -5/9
4/9 -1/9 2/9
-13/9 10/9 -11/9
Now multiply both sides of equation by this inverse matrix:
On the left side, A-1 cancel out the matrix A, and on the right side:
= -10/9 7/9 -5/9 1
4/9 -1/9 2/9 3
-13/9 10/9 -11/9 -2
= 21
-3
39
This is the solution of the system of linear equations that can be represented as:
X 21
Y = -3
Z 39
This is the second way by which any system of linear equation can be sort out but one thing to notice that every defined method for solving this type of queries required the implementation of matrices and its application, so students has to make their command on all the fundamentals and operation procedures of matrices for solving queries like system of linear equations.
So Now let us begin with the topic: operations on Polynomials;
“What is a Polynomial and how it is formed?, this query is not required to be answer for grade XI students, because you guys should aware of this, before coming into this standard. So here we are directly starting the operations discussion of polynomials:
First arithmetic operation to be applied on polynomial is Addition: while using this operation, students need to implement two standard properties that are Additional property and Distributive property that implies to combine like terms, like terms refer to same variables and same exponents for the variables.
distributive property worksheets in Algebra: if an expression is as ba + ca then it can be rewritten as a(b + c)
ba + ca = a(b + c)
let us take an example to explain it more:
Suppose given polynomial is
(3X2 + 7x + 8) + (5x2 - 8x + 2) = 3x2 + 7x + 8 + 5x2 - 8x + 2
Now use the distributive property to combine like terms as
(3x2 + 5x2) + (7x - 8x) + (8 + 2)
8x2 – x + 10
Now move towards the next operation that is Subtraction:
This one is pretty similar as Addition operation but when two terms are added then the second polynomial terms are inversed or multiplied by minus (-) sign.
We are using the same example as above:
(3X2 + 7x + 8) + [- (5x2 - 8x + 2)]
Here the second polynomial is need to be subtracted from first one that’s why firstly the second polynomial is multiplied by negative sign and then the simple addition will be performed.
3X2 + 7x + 8 - 5x2 + 8x – 2
-2x2 + 15x + 6
Third general operation, which is used to solve polynomials, is multiplication. Let us find out the simplest way to execute this operation on polynomials. While evaluating these operation students need to use above principles of addition and subtraction along with distributive property and law of exponents. Let us take an example to see the practical implementation of multiplication in between two polynomials. For more information on polynomials visit here
Example: 3x(2x -5) = (3x)(2x) – 3x(5) = 6x2 – 15x
Let us take one more example where number of derivatives is more than previous one:
( 2x + 1) (3x - 4) = (2x)(3x) – (2x)4 + (1)3x - (1)4
= 6x2 - 8x + 3x – 4
= 6x2 – 5x – 4
The fourth operation to be used to solve higher order polynomials is power operation. There are various standard formulas that are need to be learnt before starting the evaluation of complex polynomials. Some of these standard formulas are as following:
(x + y)0 = 1
(x + y)1 = x + y
(x + y)2 = x2 + 2xy + b2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
Let us take an example to use some of the above formulas for solving polynomials:
(x2 + 2x – 3)2 = (x2 + 2x – 3) (x2 + 2x – 3)
= (x2) (x2) + (x2)(2x) - (x2)(3) + (x2)(2x) + (2x)(x2) - (2x)(3) – 3(x2) - (2x)(3) +9
= x4 + 2x3 – 3x2 + 4x2 – 3x2 + 2x3 – 6x - 3x2 – 6x + 9
= x4 + 4x3 – 2x2 - 12x + 9
This is how the polynomial multiplication sorted out. While solving this student should know that whenever the bases are same, that time exponents are added to form a single term; This basic rule is learned in earlier classes but still most of the times students get confused while tracing the solved answer of the query.
Now the final operation we are going to explore in today’s article for polynomials is Division:
This operation includes some conditions that are needed to be taking in count while dividing two polynomials, as follows:
As you guys know that polynomials have many types as Monomials, Binomials and trinomials, so this is the reason the scenario of division process varies.
1. First condition is: Monomial
MonomialTo sort out this law of exponents is required:
(24x4y2)/ (18x2y3) = 4x2/3y2
This one is simple because cancellation of terms gets easier because of Monomial.
2. Polynomial
monomial:-(4x5 – 8x3 + 12x2)/ 6x2
= 4x5/6x2 – 8x3/6x2 + 12x2/6x2
= 2x3/3 – 4x/3 + 2
3. Polynomial ÷ polynomial;- for this purpose students need to follow long division algorithm that includes following steps:
a. first write the dividend and divisor polynomials into their standard polynomial form after that use zero coefficients for powers of the variable that are missing in the dividend and divisor.
b. then Divide first term of the divisor into the first terms of the dividend. Put its quotient above term in the dividend.
c. in third step, Multiply quotient by all terms of the divisor and put the products under the appropriate terms of the dividend.
d. (change signs on bottom polynomial and add) Subtract and bring down remaining terms.
e. divide the remaining term by first term until the power of the divisor is larger than the power in the dividend
Let’s take an example to execute this algorithm to solve
divide x2 + 9x +14 by (x + 7)
This is all about arithmetic operations on polynomials. If there is any doubt remaining in your mind related to polynomials operations or system of linear equation then you guys have the option of using Online math tutoring service where Online expert math tutors are waiting to give you immediate assistance with detailed explanation by solving related queries. These online tutors are available for 24 hours in a day so students can interact with them whenever they want.
In upcoming posts we will discuss about Types of Events in Grade XI and isosceles triangle theorem. Visit our website for information on Karnataka board syllabus
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