Are you preparing for your Class 9 Mathematics exam and feeling overwhelmed by Chapter 2?
In this guide, “Polynomial Maths 9 Most Important Questions with Solutions”, we have compiled the Polynomial Maths 9 Most Important Questions with step-by-step solutions. Whether you are looking for a quick revision of algebraic identities or a deep dive into cubic factorization, this “Ready-to-Post” resource has everything you need to ace your math test.
Polynomials are a fundamental part of Algebra that you will use throughout high school. Understanding how to identify types of polynomials, finding zeros, and applying theorems like the Remainder and Factor theorems is key to scoring full marks.
Polynomials are the building blocks of algebra. For Class 9 students, mastering this chapter is crucial as it forms the foundation for higher-level mathematics. This guide covers everything from basic definitions to complex factorizations.
1. Definition with Examples
A Polynomial is an algebraic expression consisting of variables, coefficients, and exponents, combined using addition, subtraction, and multiplication. The power of the variable must be a whole number (0, 1, 2, 3…).
- Example 1: 5x² + 3x + 2
- Example 2: 2y³ – 7
- Example 3: z + √2 (Note: The coefficient can be a root, but the variable z has a power of 1).
2. Why some expressions are NOT Polynomials
An expression is not a polynomial if the variable has a negative or fractional exponent.
| Expression | Reason (Why it is NOT a Polynomial) |
|---|---|
| 1/x + 2 | The variable x is in the denominator (x⁻¹), and -1 is not a whole number. |
| √x + 3 | The variable is under a root (x¹/²), and 1/2 is not a whole number. |
| y² + y³/² + 5 | The power 3/2 is a fraction, not a whole number. |
3. Types of Polynomials
A. On the basis of Number of Terms
- Monomial: Contains only one term. (e.g., 5x, 3, -7y²)
- Binomial: Contains exactly two terms. (e.g., x + 5, y² – 9)
- Trinomial: Contains exactly three terms. (e.g., x² + 2x + 1)
B. On the basis of Degree
The Degree is the highest power of the variable in the polynomial.
- Linear Polynomial: Degree is 1. (e.g., x + 3)
- Quadratic Polynomial: Degree is 2. (e.g., x² – 4)
- Cubic Polynomial: Degree is 3. (e.g., x³ + x² – 1)
- Constant Polynomial: Degree is 0. (e.g., 7, which is 7x⁰)
4. Zero of a Polynomial
A real number k is said to be a zero of a polynomial p(x) if p(k) = 0. It is the value of x that makes the whole expression zero.
Example: For p(x) = x – 5, if we put x = 5, then p(5) = 5 – 5 = 0. So, 5 is the zero of the polynomial.
5. Remainder Theorem
Definition: Let p(x) be any polynomial of degree ≥ 1 and let ‘a’ be any real number. If p(x) is divided by the linear polynomial (x – a), then the remainder is p(a).
Polynomial Maths 9 Most Important Questions: “Find the Remainder’ Type
- Q: Find the remainder when P(x)= x³ + 3x² + 3x + 1 is divided by (x + 1).
Ans: Put x = -1. then, Remainder, P(-1)=(-1)³ + 3(-1)² + 3(-1) + 1 = -1 + 3 – 3 + 1 = 0. - Q: Find the remainder when P(x)= x³ – ax² + 6x – a is divided by (x – a).
Ans: Put x = a. then, Remainder, P(a)=a³ – a(a²) + 6a – a = a³ – a³ + 5a = 5a. - Q: Find the remainder when P(x)= 4x³ – 3x² + 2x – 4 is divided by (x – 1).
Ans: p(1) = 4(1)³ – 3(1)² + 2(1) – 4 = -1. - Q: Divide p(x) = 3x⁴ – 4x³ – 3x – 1 by (x – 1).
Ans: p(1) = 3(1)⁴ – 4(1)³ – 3(1) – 1 = -5. - Q: Find remainder for x³ + 1 divided by (x + 1).
Ans: p(-1) = (-1)³ + 1 = 0.
To verify your long division and factorization results, you can use the Symbolab Polynomial Factorization Calculator.
6. Factor Theorem
Definition: (x – a) is a factor of p(x) if p(a) = 0.
Polynomial Maths 9 Most Important Questions: Check if g(x) is a factor of p(x)
- p(x) = x³ + 3x² + 5x + 6, g(x) = x + 2.
Ans: p(-2) = 0. Yes, it is a factor. - p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1.
Ans: p(-1) = 0. Yes, it is a factor. - p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2.
Ans: p(-2) = -1. No, it is not. - p(x) = x³ – 4x² + x + 6, g(x) = x – 3.
Ans: p(3) = 0. Yes, it is a factor. - p(x) = 2x⁴ + 3x³ – 2x² – 9x – 12, g(x) = x – 2.
Ans: p(2) = 18. No, it is not.
Polynomial Maths 9 Most Important Questions: Check for Multiples
- Is 2t⁴ + 3t³ – 2t² – 9t – 12 a multiple of (t – 2)?
Ans: Remainder is 18. No. - Is 7 + 3x a factor of 3x³ + 7x?
Ans: p(-7/3) ≠ 0. No. - Is x³ – x a multiple of (x – 1)?
Ans: p(1) = 0. Yes. - Is 2x + 1 a factor of 4x³ + 4x² – x – 1?
Ans: p(-1/2) = 0. Yes. - Check if x² – 1 is a multiple of x + 1.
Ans: p(-1) = 0. Yes.
7. Algebraic Identities List
-
- (a + b)² = a² + 2ab + b²
- (a – b)² = a² – 2ab + b²
- (a² – b²) = (a + b)(a – b)
- (x + a)(x + b) = x² + (a + b)x + ab
- (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 + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
- Special Case: If a + b + c = 0, then a³ + b³ + c³ = 3abc
Expanded Cubic Identities (For Quick Calculation)
- (a + b)³ = a³ + b³ + 3a²b + 3ab²
- (a – b)³ = a³ – b³ – 3a²b + 3ab²
(Note: Last term is positive)
8. Polynomial Maths 9 Most Important Questions Solved (Step-wise)
Q1. Evaluate 105 × 106 without direct multiplication.
Solution:
105 × 106 = (100 + 5)(100 + 6)
Using identity: (x+a)(x+b) = x² + (a+b)x + ab
= (100)² + (5+6)100 + (5 × 6)
= 10000 + 1100 + 30 = 11130
Q2. Factorise 49a² + 70ab + 25b².
Solution:
= (7a)² + 2(7a)(5b) + (5b)²
Using identity: x² + 2xy + y² = (x+y)²
= (7a + 5b)²
Q3. Expand (3a + 4b + 5c)².
Solution:
= (3a)² + (4b)² + (5c)² + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)
= 9a² + 16b² + 25c² + 24ab + 40bc + 30ca
Q4. Evaluate (99)³ using identity.
Solution:
99³ = (100 – 1)³
= (100)³ – (1)³ – 3(100)(1)(100 – 1)
= 1000000 – 1 – 29700 = 970299
Q5. If x + y + z = 0, find the value of x³ + y³ + z³ – 3xyz.
Solution:
We know x³+y³+z³-3xyz = (x+y+z)(x²+y²+z²-xy-yz-zx)
Since x+y+z = 0, the entire expression becomes 0.
9. All Types of Factorisation
Middle Term Splitting
Q: Factorise 6x² + 17x + 5.
= 6x² + 15x + 2x + 5
= 3x(2x + 5) + 1(2x + 5) = (3x + 1)(2x + 5)
Using Factor Theorem (Cubic)
Q: Factorise x³ – 23x² + 142x – 120.
By trial, p(1) = 0. So (x-1) is a factor. Dividing p(x) by (x-1), we get x² – 22x + 120.
Splitting middle term: (x-10)(x-12).
Final Answer: (x-1)(x-10)(x-12)
Based on Identities
Polynomial Maths 9 Most Important Questions: Cubic Factorization
Question: Factorize x³ – 23x² + 142x – 120
Step 1: Find the first factor (Trial Method)
Let p(x) = x³ – 23x² + 142x – 120.
Try x = 1:
p(1) = (1)³ – 23(1)² + 142(1) – 120
p(1) = 1 – 23 + 142 – 120 = 0
Since the result is 0, (x – 1) is a factor of the polynomial.
Step 2: Division
Divide the polynomial (x³ – 23x² + 142x – 120) by the factor (x – 1).
After division, we get the quadratic quotient:
x² – 22x + 120
Step 3: Factorize the Quadratic Quotient
Now, split the middle term of x² – 22x + 120. We need numbers that multiply to 120 and add to -22 (-12 and -10):
= x² – 12x – 10x + 120
= x(x – 12) – 10(x – 12)
= (x – 12)(x – 10)
Final Answer:
(x – 1)(x – 10)(x – 12)
For a detailed theoretical background, you can refer to the official NCERT Class 9 Mathematics Chapter 2 PDF.
Conclusion: Polynomial Class 9 Important Questions with Solutions
Mastering polynomials requires consistent practice and a clear understanding of identities. By working through these important questions, you have covered the core concepts of Class 9 Algebra—from basic definitions to the complex Factor Theorem.
Key Takeaways: Polynomial Class 9 Important Questions with Solutions
-
Always check if the power of the variable is a whole number to identify a polynomial.
-
Use the Remainder Theorem for quick calculations without long division.
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Memorize the expanded cubic identities as they are frequent in competitive exams like the UPSC CSAT or school boards.
Keep practicing, and don’t forget to bookmark this page for your final revision!