Revision Notes ofÂ Ch 2 Polynomials Class 9th Math
Topics in the Chapter- Polynomials
- Classification of polynomials on the basis of number of terms
- Degree of a Polynomial
- Values of polynomials at different points
- Zeroes of a polynomial
- Identity: (x + y + z)^{2}Â = x^{2}Â + y^{2}Â + z^{2}Â + 2xy + 2yz + 2zx
- Remainder Theorem
- Factor Theorem
- Identity: x^{3}Â + y^{3}Â + z^{3}Â â€“ 3xyz = (x + y + z)(x^{2}Â + y^{2}Â + z^{2}Â â€“ xy â€“ yz â€“ zx)
Polynomials
â†’ An algebraic expression in which the exponents of the variables are non-negative integers
are called polynomials.
For example: 3x^{4} + 2x^{3}Â + x + 9, 3x^{2}Â etc.
Constant polynomial
â†’ A constant polynomial is of the form p(x) = k, where k is a real number.
For example, â€“9, 10, 0 are constant polynomials.
Zero polynomial
A constant polynomial â€˜0â€™ is called zero polynomial.
General form of a polynomial
A polynomial of the form where are p(x) = a_{n}x^{n} + a_{n-1}x^{n-1}Â + .... + a_{1}xÂ + a_{0}x where a_{0}, a_{1}... a_{r}, are constant and a_{n}â‰ 0.
Here, a_{0}, a_{1}... a_{n}Â are the respective coefficients of x^{0},x^{1}, x^{2}Â ... x^{n} and n is the power of
the variable x.
a_{n}x^{n}Â + a_{n-1}x^{n-1}Â - a_{0} and are called the terms of p(x).
Classification of polynomials on the basis of number of terms
â€¢ A polynomial having one term is called a monomial. e.g. 3x, 25t^{3}Â etc.
â€¢ A polynomial having one term is called a monomial. e.g. 2t-6, 3x^{4}Â +2x etc.
â€¢ A polynomial having one term is called a monomial. e.g. 3x^{4}Â + 8x + 7 etc.
Degree of a Polynomial
â€¢ The degree of a polynomial is the highest exponent of the variable of the polynomial.
For example, the degree of polynomial 3x^{4}Â + 2x^{3}Â + x + 9Â is 4.
â€¢ The degree of a term of a polynomial is the value of the exponent of the term.
Classification of polynomial according to their degreesÂ
â€¢ A polynomial of degree one is called a linear polynomial e.g. 3x+2, 4x, x+9.
â€¢ A polynomial of degree is called a quadratic polynomial. e.g. x^{2}+ 9, 3x^{2}Â + 4x + 6
â€¢ A polynomial of degree three is called a cubic polynomial e.g. 10x^{3}+Â 3, 9x^{3}
Note: The degree of a non-zero constant polynomial is zero and the degree of a zero polynomial is not defined.
Values of polynomials at different points
â€¢ A polynomial is made up of constants and variables. Hence, the value of the polynomial changes
as the value of the variable in the polynomial changes.
â€¢ Thus, for the different values of the variable x, we get different values of the polynomial.
Zeroes of a polynomial
â€¢ A real number a is said to be the zero of polynomial p(x) if p(a) = 0, In this case, a is also called the root of the equation p(x) = 0.
Note:
â€¢ The maximum number of roots of a polynomial is less than or equal to the degree of the polynomial.
â€¢ A non-zero constant polynomial has no zeroes.
â€¢ A polynomial can have more than one zero.
Division of a polynomial by a monomial using long division method
Example: Divide x^{4}Â -2x^{3}- 2x^{2}+7x-15 by x-2
Identity: (x + y + z)^{2}Â = x^{2}Â + y^{2}Â + z^{2}Â + 2xy + 2yz + 2zx
Identities:Â (x + y)^{3}Â = x^{3}Â + y^{3}Â + 3xy(x+y) andÂ (x - y)^{3}Â = x^{3}Â - y^{3}Â - 3xy(x-y)
Other ways to represent these identities are:
- x^{3}+ y^{3}Â = (x+y)^{3} - 3xy(x+y)
- x^{3}+ y^{3}Â = (x+y)(x^{2}-xy+y^{2})
- x^{3}- y^{3}Â = (x-y)^{3}Â +Â 3xy(x-y)
- x^{3}- y^{3}Â = (x-y)(x^{2}+xy+y^{2})
Example: Expand (3x+2y)^{3}Â - (3x-2y)^{3}
Remainder Theorem
If p(x) is a polynomial of degree greater than or equal to one and a is any real number then if p(x) is divided by the linear polynomial x-a, the remainder is p(a).
Example:Â
Find the remainder when x^{5}Â -x^{2}Â +5 is divided by x-2.
Solution:Â
p(x) = x^{5}Â -x^{2}Â +5
The zero of x-2 is 2.
p(2) = 2^{5}Â -2^{2}Â +5 = 32-4+5 = 33
Therefore, by remainder theorem, the remainder is 33.
Factor Theorem
If p(x) is a polynomial of degree nâ‰¥1 and a is any real number, then
- x-a is a factor of p(x), if p(a)=0.Â
- p(a) = 0, if (x-a) is a factor of p(x).Â
Determine whether x +3 is a factor of x^{3}+5x^{2}+5x-3.
SolutionÂ
The zero of x+3 is -3.
Let p(x) = x^{3}+5x^{2}+5x-3
p(-3) = (3)^{3}+5(-3)^{2} + 5(-3)-3
= -27+45-15-3
= -45 + 45
= 0
Therefore, by factor theorem, x + 3 is the factor of p(x).
Factorisation of quadratic polynomials of the form ax^{2}+bx+c can be done using Factor
theorem and splitting the middle term.
Factorize x^{2}â€“ 7x + 10 using the factor theorem
Let p(x) =Â x^{2}â€“ 7x + 10Â
The constant term is 0 and its factors are Â±1, Â±2, Â±5 and Â±1O.Â
Let us check the value of the polynomial for each of these factors of 10.Â
p(1) = 1^{2}â€“ 7Ã—1 + 10 = 1-7+10 = 4 â‰ 0
Hence, x-1 is not a factor of p(x),Â
p(2) = 2^{2}â€“ 7Ã—2 + 10 = 4-14+10 = 0
Hence, x -2 is a factor of p(x).Â
p(5) = 5^{2}- 7Ã—5 +10 = 25-35+10 = 0
Hence, x -5 is a factor of p(x).Â
We know that a quadratic polynomial can have a maximum of two factors. We have obtained the two factors of the given polynomial, which are x-2 and x-5.Â
Thus, we can write the given polynomial as:
Let p(x) =Â x^{2}â€“ 7x + 10 = (x-2) (x-5)
Factorize 2x^{2}Â â€“ 11x + 15 by splitting the middle term.
The given polynomial is 2x^{2}â€“ 11x + 15
Here, a = 2Ã—15 = 30. The middle term is -11. Therefore, we have split -11 into two numbers such that their product is 30 and their sum is -11. These numbers are -5 and -6 [As (-5)+ (-6) = -11 and -5 Ã— -6 = 30]
Thus, we have:Â
2x^{2}â€“ 11x + 15 = 2x^{2}â€“ 5x - 6x + 15Â
= x(2x- 5) -3 (2x- 5)
= (x-3) (2x-5)
Note: A quadratic polynomial can have a maximum of two factors.Â
â€¢ Factorisation of cubic polynomials of the form ax^{3}+ bx^{2}+ cx can be done using factorÂ
theorem and hit and trial method.
A cubic polynomial can have a maximum of three linear factors. So, by knowing one of theseÂ
factors, we can reduce it to a quadratic polynomial.Â
Thus, to factorize a cubic polynomial, we first find a factor by the hit and trial method or by using the factor theorem, and then reduce the cubic polynomial into a quadratic polynomial and it is then solved further.Â
Factorise p(x) = x^{3}Â â€“ 7x + 6
The constant term is 6.
The factors of 6 are Â±1, Â±2, Â±3 and Â±6.
Let x = 1
From equation (1), we get
p(x) = (x â€“ 1) (x â€“ 2) (x + 3)
Hence, factors of polynomial p(x) are (x â€“ 1), (x â€“ 2) and (x + 3).
Identity: x^{3}Â + y^{3}Â + z^{3}Â â€“ 3xyz = (x + y + z)(x^{2}Â + y^{2}Â + z^{2}Â â€“ xy â€“ yz â€“ zx)