NCERT Notes for Class 9 Maths Chapter 2 Polynomials
Class 9 Maths Chapter 2 Polynomials Notes
Chapter Name | Polynomials Notes |
Class | CBSE Class 9 |
Textbook Name | Mathematics Class 9 |
Related Readings |
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Polynomials
Algebraic expression
Example: 7 + 9x – 2x2+ 5/6 xy
Constant
Example: 7, -4, 3/4 , n etc.
Variable
Example: 2x, 5x2
Terms
Example: In the expression 5x3+ 9x2+ 7x – 3, terms are 5x3, 9x2, 7x and -3.
Polynomials
Example:
(a) 5x2 + 7x + 3
(b) 9y3– 7y2 + 3y + 7
Coefficient
Note: If a term has no coefficient, the coefficient is an unwritten 1.
Example: 5x3– 7x2 – x + 3
Degree of a polynomial (in one variable)
Example: 5x + 4 is a polynomial in x of degree 1.
Degree of a polynomial in two or more variables
Example: 7x3 + 2x2y2 – 3ry + 8
Types of Polynomials
1. Linear polynomial
A polynomial of degree one is called a linear polynomial.
Example: 2x + 3 is a linear polynomial in x.
2. Quadratic polynomial
A polynomial of degree 2 is called a quadratic polynomial.
Example: 5x2 – 7x + 4 is a quadratic polynomial.
3. Cubic polynomial
A polynomial of degree 3 is called a cubic polynomial.
Example: 3x3 + 7x2 – 4x + 9 is a cubic polynomial.
4. Biquadratic polynomial
A polynomial of degree 4 is called a biquadratic polynomial.
Example: 7x4 – 2x3 + 4x + 9 is a biquadratic polynomial.
Number of Terms in a Polynomial
Categories of the polynomial according to their terms(i) Monomial
A polynomial which has only one non-zero term is called a monomial.
Example: 7, 4x, (4/5)xy, 7x2y3z5, all are monomials.
(ii) Binomial
A polynomial which has only two non-zero terms is called binomial.
Example: 2x + 7, 9x2 + 3, 3x2yz + 4x3y3z2, all are binomials.
(iii) Trinomial
A polynomial which has only three non-zero terms is called a trinomial.
Example: 5x2 + lx + 9, 5xy + 7xy2 + 3x3yz, all are trinomials.
(iv) Constant polynomial
A polynomial which has only one term and that is a constant is called a constant polynomial.
Example: (−3/4), 7, 5 all are constant polynomials.
Note: The degree of constant non-zero polynomial is zero.
(v) Zero polynomial
A polynomial which has only one term i.e., 0 is called a zero polynomial.
Note: Degree of a zero polynomial is not defined.
Value of a Polynomial
Value of a polynomial is obtained, when variable of a given polynomial is interchanged or replaced by a constant.
Let p(x) is a polynomial then value of polynomial at x = a is p(a).
Zero or root of a polynomial: A zero or root of a polynomial is the value of that variable for which value of polynomial p(x) becomes zero i.e., p(x) = 0.
Let p(x) be the polynomial and x – a.
If p(a) = 0 then real value a is called zero of a polynomial.
Remainder Theorem
Proof: Let p(x) be any polynomial of degree greater than or equal to 1. When p(x) is divided by x–a, the quotient is q(x) and remainder is r(x).
i.e. p(x) = (x-a) q(x) + r(x)
Since, degree of (x–a) is 1 and the degree of r(x) is less than the degree of (x–a) so the degree of r(x) = 0.
It: means r(x) is a constant, say r.
Therefore, for every value of x, r(x) = r
then p(x) = (x-a) q(x) + r
When x = a, then p(a) = (a – a) q(x) + r ⇒ p(a) = r
Factor Theorem
If p(x) is a polynomial of degree greater than or equal to 1 and a be any real number, then- x–a is a factor of p(x) i.e., p(x) – (x-a) q(x) which shows x–a is a factor of p(x).
- Since x–a is a factor of p(x)
p(x) = (x-a) g(x) for same polynomial g(x). In this case, p(a) = (a-a) g (a) = 0
Factorisation of the Polynomial ax2+ bx + c by Splitting the Middle Term
and factor of polynomial p(x) = (px + q) and (rx + s)
then ax2 + bx + c = (px + q) (rx + s) = prx2 + (ps +qr)x+ qs
Comparing the coefficient of x2 on both sides
a = pr ...(1)
Comparing the coefficient of x
b = ps + qr …(2)
and comparing the constant terms
c = qs …(3)
which shows that b is the sum of two numbers ps + qr.
Product of two numbers ps × qr =pr × qs = ac
So, for factors ax2 + bx + c, we should write b as sum of two numbers whose product is ac.
Here, b = p + q = 17
and ac = 6 × 5 = 30 (= pq)
then we get factors of 30, 1 × 30, 2 × 15, 3 × 10, 5 × 6,
Among above factors of 30, the sum of 2 and 15 is 17
i.e., p + q = 2 + 15 = 17
∴ 6x2 + 17x + 5
= 2x(3x + 1) + 5(3x + 1)