Tuesday, April 4, 2017

CBSE Class 9 - Maths - Chapter 2 - Polynomials (Important Points) (#cbsenotes)

Polynomials

CBSE Class 9 - Maths - Chapter 2 - Polynomials (Important Points) (#cbsenotes)
(Important Points)
CBSE Class 9 - Maths Chapter 2

A combination of constants and variables which are connected by basic mathematical operations, is known as Algebraic Expression. For e.g. x² – 7x + 2, xy² – 3 etc.

② A Polynomials p(x) in one variable x is an algebric expression in x of the form
p(x) = anxn + an–1xn–1 + ... + a1 x + a0 , where a0 , a1 , a2 ..... an are constant and a n ≠ 0 are called coefficients and n is a positive integer.

The highest power of variable x in a polynomial p(x) is called the degree of the polynomial.

A polynomial of degree 1 is called a linear polynomial. For e.g. 4x + 9 is a linear polynomial in x.



A polynomial of degree 2 is called a Quadratic Polynomial. For e.g. 7y² – 4y + 15 is a Quadratic polynomial in y.

A polynomial of degree 3 is called a Cubic Polynomial. For e.g. 3z³ – 7z²  + z – 3 is a cubic polynomial in z.


A polynomial having one term is called monomial having two terms called binomial and having three terms called trinomial.

A polynomial of degree zero is called constant polynomial.

⑨ For a polynomial p(x) if p(a) = 0 where a is a real number we say that ‘a’ is a zero of the polynomial.

⑩ If p(x) is any polynomial of degree greater than or equal to 1 and p(x) is divided by a linear polynomial x – a, then the remainder is p(a). This is called remainder theorem.

⑪ If p(x) is a polynomial of degree ≥ 1 and ‘a’ is any real number then
(x – a) is a factor of p(x), if p(a) = 0 and
p(a) = 0 if (x – a) is a factor of p(x).
This is called factor theorem.

A polynomial of degree ‘n’ can have at most n zeros.

Algebraic Identities:
(x + y)² = x² + 2xy + y²
(x – y)² = x² – 2xy + y²
x² – y² = (x – y) (x + y)
(x + y + z)²  = x²  + y² + z² + 2xy + 2yz + 2xz
(x + y)³ = x³ + y³ + 3xy (x + y)
(x – y)³  = x³ – y³ – 3xy (x – y)
x³ – y³ = (x – y) (x + xy + y )
x³ + y³ = (x + y) (x² – xy + y²)
x³ + y³ + z³ – 3xyz = (x + y + z) (x + y² + z² – xy – yz – xz)

⑭ If x + y + z = 0 then, x³ + y³ + z³ = 3xyz.


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