Polynomials
Class 9 Mathematics Chapter 2: Polynomials
Welcome to the complete, comprehensive guide to Class 9 Maths Chapter 2: Polynomials. In this article, we will explore the mathematical foundations of algebraic expressions, focusing entirely on polynomials. We will study their types, how to find their zeroes, the fascinating Remainder Theorem, factorization techniques, and standard algebraic identities. Let us dive into the world of algebra!
1. Polynomials in One Variable
Let us begin by understanding what an algebraic expression is. An algebraic expression is a combination of variables (like x, y, z) and constants (like 2, -5, 1/2) connected by mathematical operations. A Polynomial is a special type of algebraic expression.
Definition: A polynomial in one variable x is an algebraic expression of the form:
p(x) = anxn + an-1xn-1 + ... + a1x + a0
where a0, a1, ..., an are constants and the exponents of x are whole numbers.
Key Terms to Remember:
- Terms: The parts of a polynomial separated by '+' or '-' signs. For example, in
3x2 + 5x - 2, the terms are3x2,5x, and-2. - Coefficients: The numerical factor attached to the variable. In
5x, the coefficient is5. - Degree of a Polynomial: The highest power of the variable in a polynomial. For example, the degree of
4x3 - 2x2 + 8is 3.
Classification of Polynomials Based on Degree:
| Degree | Name of Polynomial | General Form | Example |
|---|---|---|---|
| 1 | Linear Polynomial | ax + b | 2x + 3 |
| 2 | Quadratic Polynomial | ax2 + bx + c | y2 - 5y + 6 |
| 3 | Cubic Polynomial | ax3 + bx2 + cx + d | 4z3 - z + 1 |
2. Zeroes of a Polynomial
When we substitute a specific real number for the variable in a polynomial, we get a value. If that specific value makes the entire polynomial equal to zero, we call that number a "Zero of the polynomial".
Definition: A real number c is a zero of the polynomial p(x) if p(c) = 0.
Numerical Example 1: Find the value of the polynomial p(x) = x2 - 3x + 2 at x = 2.
Solution:
Substitute x = 2 into the polynomial:
p(2) = (2)2 - 3(2) + 2
p(2) = 4 - 6 + 2
p(2) = 6 - 6 = 0
Since p(2) = 0, we can say that 2 is a zero of the polynomial p(x).
Important Notes on Zeroes:
- A non-zero constant polynomial (like p(x) = 5) has no zero.
- Every real number is a zero of the zero polynomial (p(x) = 0).
- A linear polynomial has exactly one zero.
- A polynomial can have more than one zero (e.g., a quadratic polynomial can have up to two zeroes).
3. Remainder Theorem
When dividing polynomials, the long division method can be tedious. The Remainder Theorem provides a brilliant shortcut to find the remainder when a polynomial is divided by a linear polynomial.
Remainder Theorem Statement: Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial (x - a), then the remainder is p(a).
Numerical Example 2: Find the remainder when p(x) = x4 + x3 - 2x2 + x + 1 is divided by (x - 1).
Solution:
Here, the divisor is x - 1. To find the root of the divisor, we set x - 1 = 0, which gives x = 1.
According to the Remainder Theorem, the remainder is p(1).
p(1) = (1)4 + (1)3 - 2(1)2 + (1) + 1
p(1) = 1 + 1 - 2 + 1 + 1
p(1) = 2 - 2 + 2 = 2
Therefore, the remainder is 2.
4. Factorisation of Polynomials
Factorisation is the reverse process of multiplication. It involves breaking down a polynomial into a product of simpler polynomials. Two primary methods are used for Class 9 Maths:
A. The Factor Theorem
The Factor Theorem is closely related to the Remainder Theorem.
Factor Theorem: If p(x) is a polynomial of degree n ≥ 1 and a is any real number, then:
(i) (x - a) is a factor of p(x), if p(a) = 0, and
(ii) p(a) = 0, if (x - a) is a factor of p(x).
B. Splitting the Middle Term
This is the most common method for factoring quadratic polynomials of the form ax2 + bx + c. We need to find two numbers p and q such that p + q = b and pq = ac.
Numerical Example 3: Factorise y2 - 5y + 6 by splitting the middle term.
Solution:
Here, a = 1, b = -5, c = 6.
We need two numbers that multiply to ac = 1 × 6 = 6 and add up to b = -5.
These two numbers are -2 and -3.
Rewrite the middle term:
y2 - 2y - 3y + 6
Take common factors from pairs:
y(y - 2) - 3(y - 2)
Take (y - 2) common:
(y - 2)(y - 3)
Hence, the factors are (y - 2) and (y - 3).
5. Algebraic Identities
An Algebraic Identity is an algebraic equation that is true for all values of the variables occurring in it. They are incredibly useful for quick calculations and factorisations. Let us deeply understand each of the standard identities for Class 9 Maths with a solved example.
- Identity I: (x + y)2 = x2 + 2xy + y2
- Identity II: (x - y)2 = x2 - 2xy + y2
- Identity III: x2 - y2 = (x + y)(x - y)
- Identity IV: (x + a)(x + b) = x2 + (a + b)x + ab
- Identity V: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
- Identity VI: (x + y)3 = x3 + y3 + 3xy(x + y)
- Identity VII: (x - y)3 = x3 - y3 - 3xy(x - y)
- Identity VIII: x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
Solved Examples for Each Identity:
Applying Identity I: (x + y)2 = x2 + 2xy + y2
Question: Expand (2a + 3b)2.
Solution: Here, x = 2a and y = 3b.
= (2a)2 + 2(2a)(3b) + (3b)2
= 4a2 + 12ab + 9b2
Applying Identity II: (x - y)2 = x2 - 2xy + y2
Question: Expand (4p - 5q)2.
Solution: Here, x = 4p and y = 5q.
= (4p)2 - 2(4p)(5q) + (5q)2
= 16p2 - 40pq + 25q2
Applying Identity III: x2 - y2 = (x + y)(x - y)
Question: Factorise 49m2 - 64n2.
Solution: This can be written as (7m)2 - (8n)2.
Here, x = 7m and y = 8n.
= (7m + 8n)(7m - 8n)
Applying Identity IV: (x + a)(x + b) = x2 + (a + b)x + ab
Question: Evaluate 105 × 106 without directly multiplying.
Solution: We can write this as (100 + 5)(100 + 6).
Here, x = 100, a = 5, and b = 6.
= (100)2 + (5 + 6)(100) + (5)(6)
= 10000 + 11(100) + 30
= 10000 + 1100 + 30 = 11130
Applying Identity V: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Question: Expand (3a + 4b + 5c)2.
Solution: Here, x = 3a, y = 4b, and z = 5c.
= (3a)2 + (4b)2 + (5c)2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)
= 9a2 + 16b2 + 25c2 + 24ab + 40bc + 30ac
Applying Identity VI: (x + y)3 = x3 + y3 + 3xy(x + y)
Question: Expand (2x + 3)3.
Solution: Here, x = 2x and y = 3.
= (2x)3 + (3)3 + 3(2x)(3)(2x + 3)
= 8x3 + 27 + 18x(2x + 3)
= 8x3 + 36x2 + 54x + 27
Applying Identity VII: (x - y)3 = x3 - y3 - 3xy(x - y)
Question: Evaluate (99)3 using a suitable identity.
Solution: Write 99 as (100 - 1).
= (100 - 1)3
= (100)3 - (1)3 - 3(100)(1)(100 - 1)
= 1000000 - 1 - 300(99)
= 1000000 - 1 - 29700
= 970299
Applying Identity VIII: x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
Question: Factorise 8x3 + y3 + 27z3 - 18xyz.
Solution: First, rewrite it in the form of cubes:
= (2x)3 + (y)3 + (3z)3 - 3(2x)(y)(3z)
Here, replace x with 2x, y with y, and z with 3z in the identity.
= (2x + y + 3z)[(2x)2 + y2 + (3z)2 - (2x)(y) - (y)(3z) - (3z)(2x)]
= (2x + y + 3z)(4x2 + y2 + 9z2 - 2xy - 3yz - 6zx)
Conclusion
Mastering Class 9 Maths Chapter 2 Polynomials is essential, as these concepts form the foundation for higher-level mathematics. By understanding polynomials in one variable, zeroes, the remainder theorem, factorisation, and practically applying algebraic identities, solving complex algebra problems becomes much easier and highly logical.
Comments
Post a Comment