The CBSE Notes for Class 10 Maths for Chapter 2 Polynomials help students learn about polynomials and become proficient in algebraic expressions. The most recent CBSE curriculum, which serves as the basis for these lectures, breaks down each subject into easily understood principles, formulas, and examples for practice on tests. Students are prepared for more complex algebraic topics in the higher courses by gaining a fundamental grasp of polynomials, types, and operations in Chapter 2.
Students can download CBSE Notes for Class 10 free PDF notes for Maths second chapter, Polynomials. This readily accessible resource is designed to facilitate easy understanding of the concepts within the Polynomials chapter.
This chapter introduces the idea of polynomials. Expressions that contain exponents, constants, and variables are called polynomials. In algebra, it is a crucial issue. The following lists some of the key ideas that lie under the umbrella of "polynomials":
Definition of Polynomials: Expressions that include variables with non-negative integer exponents, such as
For ax2 + bx + c: Sum of zeros =
Product of zeros =
Degree of Polynomial
Types of Polynomials
ON BASIS OF TERMS
Monomial (2x)
Binomial (x² + 3x)
Trinomial (3x³ − x² + 3x)
Quadrinomial (2x⁴ + 3x³ − x² + 3x)
ON BASIS OF DEGREE
Zero (0)
Constant (3)
Linear (2x + 1)
Quadratic (x² − 3)
Cubic (x³ − 4x + 3)
Bi-Quadratic (x⁴ − x + 5)
Zero or Root of polynomial
Sol.
p(2) = (2)² − 5 × 2 + 6 = 4 − 10 + 6 = 0
p(3) = (3)² − 5 × 3 + 6 = 9 − 15 + 6 = 0
∴ 2 and 3 are the zeros of the polynomial p(x).
Value of Polynomial
The value of a polynomial f(x) at x = a is obtained by substituting x = a in given polynomial.
For e.g. Find the value of polynomial 5x − 4x² + 3 at x = −1.
Sol. p(x) = 5x − 4x² + 3
To find value of polynomial at x = −1 or p(−1)
So we put x = −1 in polynomial,
p(−1) = 5 × (−1) − 4 × (−1)² + 3
p(−1) = −5 − 4 + 3
p(−1) = −6
So value of polynomial at x = −1 is −6.
If p(x) is divided by (x − a) then remainder is given by p(a).
For e.g. Find the remainder when polynomial p(x) = x³ − 4x² − 7x + 10 is divided by x − 2.
Sol.
p(x) = x³ − 4x² − 7x + 10
p(2) = 2³ − 4(2)² − 7(2) + 10
= 8 − 16 − 14 + 10 = −12
Factor Theorem
Sol. We know that the zero of the polynomial (x + 2) is −2.
Let p(x) = x³ + 3x² + 5x + 6
Then,
p(−2) = (−2)³ + 3(−2)² + 5(−2) + 6
= −8 + 12 − 10 + 6 = 0
Hence x + 2 is factor of given polynomial.
Geometrical meaning of the zeros of a polynomial
Case-I : Two distinct zeros
Case-II : One zero
Case-III : No zero
Zero of a linear polynomial = − b / a= - (constant term / coefficient of x)
Then polynomial f(x) is given by f(x) = K{x² − (α + β)x + αβ}
Sum of zeros = α + β + γ = − b / a = - (coefficient of x² / coefficient of x³)
Sum of product of 2 zeroes at a time = αβ + βγ + γα = c / a = (coefficient of x / coefficient of x³)
Product of zeros = αβγ = − d / a = (constant term / coefficient of x³)
Polynomial f(x) is given by f(x) = K [x³ − (α + β + γ)x² + (αβ + βγ + γα)x − αβγ]
If f(x) is a polynomial and g(x) is a non-zero polynomial, then there exist two polynomials q(x) and r(x) such that f(x) = g(x) × q(x) + r(x), where r(x) = 0 or degree r(x) < degree g(x). In other words,
Dividend = Divisor × Quotient + Remainder
Remark: If r(x) = 0, then polynomial g(x) is a factor of polynomial f(x).
Some useful relations involving α and β are
α² + b² = (α + b)² − 2αb
(α − b)² = (α + b)² − 4αb
α² − b² = (α + b)(α − b) = (α + b) √((α + b)² − 4αb)
α³ + b³ = (α + b)³ − 3αb (α + b)
α³ − b³ = (α − b)³ + 3αb (α − b)
α⁴ − b⁴ = (α² + b²) (α + b) (α − b)= [(α + b)² − 2αb] (α + b) √((α + b)² − 4αb)
α⁴ + b⁴ = (α² + b²)² − 2(αb)²= [(α + b)² − 2αb]² − 2(αb)²
α⁵ + b⁵ = (α³ + b³) (α² + b²) − α²b² (α + b) = [(α + b)³ − 3αb (α + b)] [(α + b)² − 2αb] − (αb)² (α + b)
(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
a² − b² = (a + b)(a − b)
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
(a + b)³ = a³ + b³ + 3ab(a + b)
(a − b)³ = a³ − b³ − 3ab(a − b)
a³ + b³ = (a + b)(a² − ab + b²)
a³ − b³ = (a − b)(a² + ab + b²)
a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
Special case: If a + b + c = 0 then a³ + b³ + c³ = 3abc
For students looking to improve their algebraic skills, these CBSE Maths Notes for Class 10 Chapter 2 on Polynomials are a great resource. By emphasising terminology, formulae, and solved problems, the notes help students become more comfortable answering questions pertaining to polynomials on tests.
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