Polynomial Expression:
A polynomial expression is a mathematical expression consisting of one or more variables and constants, and where the powers of the variables are positive integers. For example, x, x2, x3 etc.
Polynomial expressions are usually written as:
P(x) = anxn + an-1xn-1 +…..+ a1x + a0
here,
an,an-1,....,a1,a0 are constants.
x is variable.
n is the highest degree of the polynomial.
Example:
Let a polynomial expression be
P(x) = anxn + an-1xn-1 +…..+ a1x + a0
here,
an,an-1,....,a1,a0 are constants.
x is variable.
n is the highest degree of the polynomial.
Example:
Let a polynomial expression be
P(x) = 2x3 + 3x2 - 5x + 7
Now decomposing this polynomial expression is:
2x3: Here 2 is the constant, and x3 is the power of the variable.
3x2: Here 3 is the constant, and x2 is the power of the variable.
-5x: Here -5 is the constant, and x is the power of the variable.
7: It is only constant term.
The highest power of this polynomial is 3, so the degree of this polynomial is 3.
Name of parts:
2x3: Term to the third power
3x2: Term of the second power
-5x: First power term
7: Constant term
Use of polynomial expressions:
Examples of polynomial expressions:
Let a polynomial expression
Now decomposing this polynomial expression is:
2x3: Here 2 is the constant, and x3 is the power of the variable.
3x2: Here 3 is the constant, and x2 is the power of the variable.
-5x: Here -5 is the constant, and x is the power of the variable.
7: It is only constant term.
The highest power of this polynomial is 3, so the degree of this polynomial is 3.
Name of parts:
2x3: Term to the third power
3x2: Term of the second power
-5x: First power term
7: Constant term
Use of polynomial expressions:
Polynomials are used in mathematical analysis, physics, engineering, economics, and many other fields. Such expressions play an important role in solving equations, graphing functions and solving various mathematical problems.
Examples of polynomial expressions:
Let a polynomial expression
P(x) = 3x4 - 2x3 + 5x2 - 7x + 4
1. Constant
1. Constant
A constant is a number that is not related to the variable in a polynomial expression and does not change.
The constant in this polynomial is 4.
2. Term
A term is each separate part of a polynomial consisting of variables and coefficients.
P(x) = 3x4 - 2x3 + 5x2 - 7x + 4 ; The number of terms in this polynomial is five:
3x4
-2x3
5x2
-7x
4
3. Degree
The magnitude or power of a term is the power or exponent of the variable.
P(x) = 3x4 - 2x3 + 5x2 - 7x + 4
Here are the degree of each term:
3x4 terms have degree 4
-2x3 term has degree 3
5x2 terms have degree 2
-7x terms of degree 1
4 terms of degree 0 (the term has no exponent, it is just a constant)
4. Constant Term
A constant is a term of a polynomial that has no variable, i.e. it is only a constant.
P(x) = 3x4 - 2x3 + 5x2 - 7x + 4
Here the constant is 4
5. Degree of Polynomial
P(x) = 3x4 - 2x3 + 5x2 - 7x + 4 ;
5. Degree of Polynomial
The highest degree of a polynomial is the degree of the term with the highest power.
P(x) = 3x4 - 2x3 + 5x2 - 7x + 4 ;
The highest degree of this polynomial is 4, because the power of the 3x4 term is the highest.
6. Coefficient
P(x) = 3x4 - 2x3 + 5x2 - 7x + 4
Here are the coefficients of each term:
The coefficient of 3x4 term is 3
The coefficient of the -2x3 term is -2
The coefficient of 5x2 term is 5
The coefficient of the -7x term is -7
Coefficient of 4 terms is 4
Another example:
Let another polynomial Q(x) = x3 + 4x2 - x + 6
here,
The constant is 6
Terms: x3, 4x2, -x, 6
Degree:
The term x3 has degree 3
4x2 terms have degree 2
The degree of the -x term is 1
6 terms have degree 0
Constant: 6
Degree of polynomial: 3 (highest power term x3)
Coefficient:
The coefficient of the x3 term is 1
The coefficient of the term 4x2 is 4
The coefficient of the -x term is -1
Coefficient of 6 terms is 6
Monomial Expressions with One Variable
Example: 5x3
Here:
Terms: 5x3
Variable: x
Degree: 3 (power of variable)
Coefficient: 5
Another example might be: -2x
Here:
Terms: -2x
Variable: x
Degree: 1 (exponent of variable)
Coefficient: -2
Binomial Expressions with One Variable
Example: 3x2 + 4x
Here:
First term: 3x2
Second term: 4x
Variable: x
Degree of first term: 2 (power of variable)
Degree of second term: 1 (power of variable)
Coefficient of first term: 3
Coefficient of second term: 4
Polynomial Expression with Two Variables
Example:
P(x,y) = 3x2 + 2xy + y2
Here:
3x2 is the term of x to the degree 2.
2xy is the product of x and y.
y2 is the degree 2 of y.
Polynomials of Three Variables
Example:
Q(x,y,z) = x2 + y2 + z2 + 2xy + 2yz + 2zx
Here:
x2, y2, z2 are the quadratic terms of x, y, and z.
2xy, 2yz, 2zx are the product of variables.
Leading Term
In a polynomial expression, the leading term is the term with the highest coefficient of the variable.
Example: 4x5 + 3x3 - 2x + 7
Here:
Key terms: 4x5 (because its degree or power is highest, ie 5)
Leading Coefficient
The principal coefficient is the coefficient of the principal term.
Example: 4x5 + 3x3 - 2x + 7
Here:
Keywords: 4x5
Leading Coefficient: 4
Homogeneous Polynomial
Homogeneous polynomials are polynomials in which each term has the same degree.
Example: P(x,y) = 3x2y + 4xy2
Here the total degree of each term is 3:
The total degree of 3x2y terms is 2 + 1 = 3
Total degree of 4xy2 terms is 1 + 2 = 3
Then, P(x,y) is an homogeneous polynomial, since the sum of all terms is the same, i.e. 3.
Another example: Q(a,b,c) = 2a3 + 3b3 + 4c3
Here the total degree of each term is 3:
2a3 terms sum to degree 3
The total degree of 3b3 terms is 3
4c3 terms have a total degree of 3
Then, Q(a,b,c) is a homogeneous polynomial.
Symmetric Polynomial
A symmetric polynomial is a polynomial that remains the same if the order of its variables is changed.
Example: P(x,y) = x2 + y2 + xy
Here P(x,y) and P(y,x) are same, ie:
P(x,y) = x2 + y2 + xy
P(y,x) = y2 + x2 + yx
Both are identical polynomials, so P(x,y) is a symmetric polynomial.
Another example: Q(a,b,c) = a2 + b2 + c2 + ab + bc + ca
Here, Q(a,b,c), Q(b,c,a), Q(c,a,b) are the same, so Q(a,b,c) is a symmetric polynomial.
Cyclic Polynomial
Cyclic polynomials are polynomials that remain identical when the variables are rotated clockwise or counterclockwise.
Example: P(x,y,z) = x2y + y2z + z2x
Here P(x,y,z), P(y,z,x) and P(z,x,y) are same, ie:
P(x,y,z) = x2y + y2z + z2x
P(y,z,x) = y2z + z2x + x2y
P(z,x,y) = z2x + x2y + y2z
Then, P(x,y,z) is a cyclic polynomial.
Addition of Polynomials
Addition of Polynomials, terms of the same degree must be combined.
Example: (3x2 + 2x + 1) + (4x2 - 3x + 5)
Solution:
= 3x2 + 4x2 + 2x - 3x + 1 + 5
= 7x2 - x + 6
Subtraction of Polynomials
Subtraction of Polynomials, terms of the same power must be subtracted from each other.
Example: (5x2 + 3x - 2) - (2x2 - 4x + 1)
Solution:
= 5x2 - 2x2 + 3x - (- 4x) - 2 - 1
= 3x2 + 7x - 3
Multiplication of Polynomials
Multiplying polynomials involves multiplying each term and adding the exponents.
Example: (2x + 3)⋅(x + 4)
Solution:
= 2x⋅x + 2x⋅4 + 3⋅x + 3⋅4
= 2x2 + 8x + 3x + 12
= 2x2 + 11x + 12
Division of Polynomials
Step 1: Determine the first term
First, work with highest power terms.
1. The first term is 2x3 and the quotient is 2x3 x = 2x2
2. Then multiply 2x2 by x - 1 and subtract 2x3 + 3x2 - x + 5.
Step 2: Finding the second term
Now divide the term 5x2 by x.
1. The second term is 5x2 and the quotient is 5x2 x = 5x
2. Then multiply 5x by x - 1 and subtract 5x2 - x + 5.
Step 3: Finding the third term
Now divide the term 4x by x.
1. The third term is 4x and the quotient will be 4x x = 4
2. Then multiply 4 by x - 1 and subtract from 4x + 5.
Final result
So, the quotient is 2x2 + 5x + 4 and the quotient is 9
So: 2x3 + 3x2 - x + 5 x - 1 = 2x2 + 5x + 4 + 9x - 1
Remainder Theorem
Explain with example:
Let's say, p(x) = x3 - 4x2 + x + 6 and we need to divide p(x) by x - 2.
1. First find the value of p(a) by substituting the value of a into p(x):
If a = 2:
p(2) = 23 - 4 ✕ 22 + 2 + 6
= 8 - 16 + 2 + 6
= 0
Then the quotient will be p(2) = 0
2. Another example:
p(x) = x2 + 3x + 2 and a = -1
p(-1) = (-1)2 + 3 ✕ (-1) + 2
= 1 - 3 + 2
= 0
Then the quotient will be p(-1) = 0
In summary: if p(x) - is divided by (x - a), then the quotient is p(a).
Factor Theorem
Explain with example:
Let p(x) = x2 - 5x + 6
1. First check the value of a:
If a = 2:
p(2) = 22 - 5 ✕ 2 + 6 = 4 - 10 + 6 = 0
Here we see that p(2) = 0, so (x - 2) is a factor of p(x).
2. Let's check another value of a:
If a = 3:
p(3) = 32 - 5 ✕ 3 + 6 = 9 - 15 + 6 = 0
Here we see that p(3) = 0, so (x - 3) is a factor of p(x) - .
In summary: if p(a) = 0, then (x - a) is a factor of p(x) - .
Perfect Square Expression
General form of square root:
1. (a + b)2 = a2 + 2ab + b2
2. (a - b)2 = a2 - 2ab + b2
Explain with example:
1. x2 + 6x + 9
First, let's check if it's a perfect square expression:
Here a = x and b = 3
Checking in the form a2 + 2ab + b2:
= x2 + 2 ✕ x ✕ 3 + 32
= x2 + 6x + 9
So, x2 + 6x + 9 is a perfect square expression and can be written in the form (x + 3)2.
2. 4x2 - 12x + 9
First, let's check if it's a perfect square expression:
Here a = 2 and b = 3
Checking in the form a2 - 2ab + b2:
= (2x)2 - 2 ✕ 2x ✕ 3 + 32
= 4x2 - 12x + 9
So, 4x2 - 12x + 9 is a perfect square and can be written as (2x - 3)2.
Summary: A quadratic expression is a polynomial expression that can be written as a perfect square expression.
Example: x2 + 5x + 6
Solution: (x + 2)(x + 3)
Factorization through Splitting of the Middle Term
Example: x2 + 7x + 12
Solution:
= x2 + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 3)(x + 4)
Factorization in general way
x = -b ±√ b2 − 4ac 2a
Here, two roots or values of x are found x1 and x2; b2−4ac is called the "judge" or "divisor".
If b2−4ac＞0, then there are two real and distinct roots.
If b2−4ac＞0, then there is a double real root.
If b2−4ac＞0, then there are no real roots, two imaginary (complex) roots.
Identity:
Explain with example:
1. Identity Examples:
Let p(x) = (x+1)2 and q(x) = x2 + 2x + 1
Expanding p(x) = (x + 1)2 here gives:
(x + 1)2 = x2 + 2x + 1
So, p(x)≡q(x) because p(x) = q(x) is true for all x.
In summary:
Identity is an equation that is true for all x.
This is written p(x)≡q(x)
Partial Fraction
Types of Partial Fractions:
Steps in Partial Fractions:
1. Division into proper fractions:
Let's say, 3x + 1 (x - 1)(x + 2) is a proper fraction.
We can divide this fraction into partial fractions:
3x + 1(x - 1)(x + 2) = Ax - 1 + Bx + 2
where a and b are constants. To determine a and b:
3x + 1 = A(x + 2) + B(x - 1)
3x + 1 = Ax + 2A + Bx - B
3x + 1 = (A + B)x + (2A - B)
Hence, a and b must be determined by solving A + B = 3 and 2A - B = 1.
Explanation with Example: proper fractions
Let's say, 3x + 1 (x - 1)(x + 2)
3x + 1(x - 1)(x + 2) = A(x - 1) + B(x + 2)
Determine a and b:
If A = 1 and B = 2:
3x + 1(x - 1)(x + 2) = 1(x - 1) + 2 (x + 2)
2. Division into Improper Fractions:
Let's say, x3 + 1x2 + 1 is an improper fraction.
Methods of Converting Improper Fractions
First split, x3 + 1x2 + 1 = x + (-x + 1) (x2 + 1) must be divided into partial fractions.
Explain with example:
Let's say, x3 + 1x2 + 1
First divide by:
x3 + 1x2 + 1 = x + (-x + 1) (x2 + 1)
Here x is the quotient and - x + 1x2 + 1 is a partial fraction.
In summary:
Partial fractions are the process of dividing a rational fraction into smaller fractions.
proper fractions are divided into partial fractions by determining the quotient.
6. Coefficient
The coefficient is the numerical value of the term multiplied by the variable.
P(x) = 3x4 - 2x3 + 5x2 - 7x + 4
Here are the coefficients of each term:
The coefficient of 3x4 term is 3
The coefficient of the -2x3 term is -2
The coefficient of 5x2 term is 5
The coefficient of the -7x term is -7
Coefficient of 4 terms is 4
Another example:
Let another polynomial Q(x) = x3 + 4x2 - x + 6
here,
The constant is 6
Terms: x3, 4x2, -x, 6
Degree:
The term x3 has degree 3
4x2 terms have degree 2
The degree of the -x term is 1
6 terms have degree 0
Constant: 6
Degree of polynomial: 3 (highest power term x3)
Coefficient:
The coefficient of the x3 term is 1
The coefficient of the term 4x2 is 4
The coefficient of the -x term is -1
Coefficient of 6 terms is 6
Monomial Expressions with One Variable
A monomial expression with one variable is a mathematical expression that consists of only one term and has one variable.
Example: 5x3
Here:
Terms: 5x3
Variable: x
Degree: 3 (power of variable)
Coefficient: 5
Another example might be: -2x
Here:
Terms: -2x
Variable: x
Degree: 1 (exponent of variable)
Coefficient: -2
Binomial Expressions with One Variable
A binomial expression with one variable is a mathematical expression consisting of two terms and one variable.
Example: 3x2 + 4x
Here:
First term: 3x2
Second term: 4x
Variable: x
Degree of first term: 2 (power of variable)
Degree of second term: 1 (power of variable)
Coefficient of first term: 3
Coefficient of second term: 4
Polynomial Expression with Two Variables
An example of a polynomial expression in two variables is a polynomial expression that consists of two variables.
Example:
P(x,y) = 3x2 + 2xy + y2
Here:
3x2 is the term of x to the degree 2.
2xy is the product of x and y.
y2 is the degree 2 of y.
Polynomials of Three Variables
An example of a polynomial expression in three variables is a polynomial expression that consists of three variables.
Example:
Q(x,y,z) = x2 + y2 + z2 + 2xy + 2yz + 2zx
Here:
x2, y2, z2 are the quadratic terms of x, y, and z.
2xy, 2yz, 2zx are the product of variables.
Leading Term
In a polynomial expression, the leading term is the term with the highest coefficient of the variable.
Example: 4x5 + 3x3 - 2x + 7
Here:
Key terms: 4x5 (because its degree or power is highest, ie 5)
Leading Coefficient
The principal coefficient is the coefficient of the principal term.
Example: 4x5 + 3x3 - 2x + 7
Here:
Keywords: 4x5
Leading Coefficient: 4
Homogeneous Polynomial
Homogeneous polynomials are polynomials in which each term has the same degree.
Example: P(x,y) = 3x2y + 4xy2
Here the total degree of each term is 3:
The total degree of 3x2y terms is 2 + 1 = 3
Total degree of 4xy2 terms is 1 + 2 = 3
Then, P(x,y) is an homogeneous polynomial, since the sum of all terms is the same, i.e. 3.
Another example: Q(a,b,c) = 2a3 + 3b3 + 4c3
Here the total degree of each term is 3:
2a3 terms sum to degree 3
The total degree of 3b3 terms is 3
4c3 terms have a total degree of 3
Then, Q(a,b,c) is a homogeneous polynomial.
Symmetric Polynomial
A symmetric polynomial is a polynomial that remains the same if the order of its variables is changed.
Example: P(x,y) = x2 + y2 + xy
Here P(x,y) and P(y,x) are same, ie:
P(x,y) = x2 + y2 + xy
P(y,x) = y2 + x2 + yx
Both are identical polynomials, so P(x,y) is a symmetric polynomial.
Another example: Q(a,b,c) = a2 + b2 + c2 + ab + bc + ca
Here, Q(a,b,c), Q(b,c,a), Q(c,a,b) are the same, so Q(a,b,c) is a symmetric polynomial.
Cyclic Polynomial
Cyclic polynomials are polynomials that remain identical when the variables are rotated clockwise or counterclockwise.
Example: P(x,y,z) = x2y + y2z + z2x
Here P(x,y,z), P(y,z,x) and P(z,x,y) are same, ie:
P(x,y,z) = x2y + y2z + z2x
P(y,z,x) = y2z + z2x + x2y
P(z,x,y) = z2x + x2y + y2z
Then, P(x,y,z) is a cyclic polynomial.
Addition of Polynomials
Addition of Polynomials, terms of the same degree must be combined.
Example: (3x2 + 2x + 1) + (4x2 - 3x + 5)
Solution:
= 3x2 + 4x2 + 2x - 3x + 1 + 5
= 7x2 - x + 6
Subtraction of Polynomials
Subtraction of Polynomials, terms of the same power must be subtracted from each other.
Example: (5x2 + 3x - 2) - (2x2 - 4x + 1)
Solution:
= 5x2 - 2x2 + 3x - (- 4x) - 2 - 1
= 3x2 + 7x - 3
Multiplication of Polynomials
Multiplying polynomials involves multiplying each term and adding the exponents.
Example: (2x + 3)⋅(x + 4)
Solution:
= 2x⋅x + 2x⋅4 + 3⋅x + 3⋅4
= 2x2 + 8x + 3x + 12
= 2x2 + 11x + 12
Division of Polynomials
To divide a polynomial, the quotient and quotient must be determined.
A simple step-by-step procedure for dividing one polynomial by another polynomial will be shown. Here we have 2x3 + 3x2 - x + 5 divided by x - 1. For this, polynomial long division will be used.
Step 1: Determine the first term
First, work with highest power terms.
1. The first term is 2x3 and the quotient is 2x3 x = 2x2
2. Then multiply 2x2 by x - 1 and subtract 2x3 + 3x2 - x + 5.
2x2 × (x - 1) = 2x3 - 2x2
(2x3 + 3x2 - x + 5) - (2x3 - 2x2) = 3x2 - (- 2x2) - x + 5 = 5x2 - x + 5
Step 2: Finding the second term
Now divide the term 5x2 by x.
1. The second term is 5x2 and the quotient is 5x2 x = 5x
2. Then multiply 5x by x - 1 and subtract 5x2 - x + 5.
5x × (x - 1) = 5x2 - 5x
(5x2 - x + 5) - (5x2 - 5x) = - x - (-5x) + 5 = 4x + 5
Step 3: Finding the third term
Now divide the term 4x by x.
1. The third term is 4x and the quotient will be 4x x = 4
2. Then multiply 4 by x - 1 and subtract from 4x + 5.
4 × (x - 1) = 4x - 4
(4x + 5) - (4x - 4) = 5 - (- 4) = 5 + 4 = 9
Final result
So, the quotient is 2x2 + 5x + 4 and the quotient is 9
So: 2x3 + 3x2 - x + 5 x - 1 = 2x2 + 5x + 4 + 9x - 1
Remainder Theorem
If p(x) is a polynomial and a is a constant, then dividing p(x) by x - a yields p(a).
Explanation: The quotient theorem tells us that if we divide p(x) by x - a, the quotient will be the value of p(a). That is, the value of p(x) obtained by substituting a for x will be the last division.
Explain with example:
Let's say, p(x) = x3 - 4x2 + x + 6 and we need to divide p(x) by x - 2.
1. First find the value of p(a) by substituting the value of a into p(x):
If a = 2:
p(2) = 23 - 4 ✕ 22 + 2 + 6
= 8 - 16 + 2 + 6
= 0
Then the quotient will be p(2) = 0
2. Another example:
p(x) = x2 + 3x + 2 and a = -1
p(-1) = (-1)2 + 3 ✕ (-1) + 2
= 1 - 3 + 2
= 0
Then the quotient will be p(-1) = 0
In summary: if p(x) - is divided by (x - a), then the quotient is p(a).
Factor Theorem
If p(x) is a polynomial and a is a constant, then (x - a) is a derivative of p(x) - if and only if p(a) = 0.
Explanation: The factor theorem tells us that (x - a) of p(x) has a factor only if p(a) has a value of 0. That is, if p(x) has value 0 by substituting a for x, then x - a is a factor of p(x).
Explain with example:
Let p(x) = x2 - 5x + 6
1. First check the value of a:
If a = 2:
p(2) = 22 - 5 ✕ 2 + 6 = 4 - 10 + 6 = 0
Here we see that p(2) = 0, so (x - 2) is a factor of p(x).
2. Let's check another value of a:
If a = 3:
p(3) = 32 - 5 ✕ 3 + 6 = 9 - 15 + 6 = 0
Here we see that p(3) = 0, so (x - 3) is a factor of p(x) - .
In summary: if p(a) = 0, then (x - a) is a factor of p(x) - .
Perfect Square Expression
A perfect square expression is a polynomial expression that can be written in the form of a square expression. That is, if a polynomial expression can be written as (a + b)2 or (a - b)2, then it is called a perfect square expression.
General form of square root:
1. (a + b)2 = a2 + 2ab + b2
2. (a - b)2 = a2 - 2ab + b2
Explain with example:
1. x2 + 6x + 9
First, let's check if it's a perfect square expression:
Here a = x and b = 3
Checking in the form a2 + 2ab + b2:
= x2 + 2 ✕ x ✕ 3 + 32
= x2 + 6x + 9
So, x2 + 6x + 9 is a perfect square expression and can be written in the form (x + 3)2.
2. 4x2 - 12x + 9
First, let's check if it's a perfect square expression:
Here a = 2 and b = 3
Checking in the form a2 - 2ab + b2:
= (2x)2 - 2 ✕ 2x ✕ 3 + 32
= 4x2 - 12x + 9
So, 4x2 - 12x + 9 is a perfect square and can be written as (2x - 3)2.
Summary: A quadratic expression is a polynomial expression that can be written as a perfect square expression.
Factorization of Quadratic Expressions
To factorization of Quadratic Expressions, ax2 + bx + c divides the two factors.
Example: x2 + 5x + 6
Solution: (x + 2)(x + 3)
Factorization through Splitting of the Middle Term
To find the product in the factorization method, the product is divided into two terms and then their common product is found.
Example: x2 + 7x + 12
Solution:
= x2 + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 3)(x + 4)
Factorization in general way
A quadratic equation is a polynomial equation whose highest degree is two. This is usually written as: ax2+bx+c = 0 where a, b, and c are constants, and a≠0
x = -b ±√ b2 − 4ac 2a
Here, two roots or values of x are found x1 and x2; b2−4ac is called the "judge" or "divisor".
If b2−4ac＞0, then there are two real and distinct roots.
If b2−4ac＞0, then there is a double real root.
If b2−4ac＞0, then there are no real roots, two imaginary (complex) roots.
Identity:
Identity is an equation that is true for all values of x. This means, any value of x in a differential equation will be correct. Identity is expressed with the symbol ≡.
General Form of Identity: If p(x) and q(x) are two polynomials and p(x) = q(x) for all x, then p(x) and q(x) are differentiable. It is written: p(x)≡q(x)
Explain with example:
1. Identity Examples:
Let p(x) = (x+1)2 and q(x) = x2 + 2x + 1
Expanding p(x) = (x + 1)2 here gives:
(x + 1)2 = x2 + 2x + 1
So, p(x)≡q(x) because p(x) = q(x) is true for all x.
In summary:
Identity is an equation that is true for all x.
This is written p(x)≡q(x)
Partial Fraction
Partial fractions are the process by which a rational fraction is divided into several smaller fractions. It is helpful to easily analyze polynomial fractions and solve equations.
Types of Partial Fractions:
1. Proper Fraction: When the magnitude of the product of the polynomial is smaller than the degree of the determinant.
2. Improper Fraction: When the magnitude of the product of the polynomial is equal to or greater than the degree of the determinant.
Steps in Partial Fractions:
1. Division into proper fractions:
Let's say, 3x + 1 (x - 1)(x + 2) is a proper fraction.
We can divide this fraction into partial fractions:
3x + 1(x - 1)(x + 2) = Ax - 1 + Bx + 2
where a and b are constants. To determine a and b:
3x + 1 = A(x + 2) + B(x - 1)
3x + 1 = Ax + 2A + Bx - B
3x + 1 = (A + B)x + (2A - B)
Hence, a and b must be determined by solving A + B = 3 and 2A - B = 1.
Explanation with Example: proper fractions
Let's say, 3x + 1 (x - 1)(x + 2)
3x + 1(x - 1)(x + 2) = A(x - 1) + B(x + 2)
Determine a and b:
If A = 1 and B = 2:
3x + 1(x - 1)(x + 2) = 1(x - 1) + 2 (x + 2)
2. Division into Improper Fractions:
Let's say, x3 + 1x2 + 1 is an improper fraction.
Methods of Converting Improper Fractions
An improper fraction is a fraction that has a large numerator and a small denominator. To convert an improper fraction to a proper fraction, express it as the sum of the whole number and the fraction. This will be clear from the following examples.
First split, x3 + 1x2 + 1 = x + (-x + 1) (x2 + 1) must be divided into partial fractions.
Explain with example:
Let's say, x3 + 1x2 + 1
First divide by:
x3 + 1x2 + 1 = x + (-x + 1) (x2 + 1)
Here x is the quotient and - x + 1x2 + 1 is a partial fraction.
In summary:
Partial fractions are the process of dividing a rational fraction into smaller fractions.
proper fractions are divided into partial fractions by determining the quotient.
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