NCERT Solutions for Class 9 Maths Exercise 2.2 Question 2
This guide explains how to find the value of a polynomial for different inputs. We will solve all four subparts of Question 2 from Exercise 2.2 by substituting values into the given algebraic expressions.
Question 2: Find &&p(0)&&, &&p(1)&& and &&p(2)&& for each of the following polynomials.
How to Find the Value of a Polynomial
- Identify the Polynomial and Value: Identify the given polynomial, for example, &&p(y) = y^2 – y + 1&&, and the value to be substituted for the variable, for example, &&y = 0&&.
- Substitute the Value: Replace every instance of the variable in the polynomial with the given numerical value. Use parentheses to prevent sign errors.
- Simplify the Expression: Simplify the resulting expression using the correct order of operations (PEMDAS/BODMAS) to find the final numerical answer.
Step-by-Step Solutions 📝
(i) For polynomial &&p(y) = y^2 – y + 1&&
- For &&p(0)&&: Substitute &&y=0&&.
&&p(0) = (0)^2 – (0) + 1 = 0 – 0 + 1 = 1&& - For &&p(1)&&: Substitute &&y=1&&.
&&p(1) = (1)^2 – (1) + 1 = 1 – 1 + 1 = 1&& - For &&p(2)&&: Substitute &&y=2&&.
&&p(2) = (2)^2 – (2) + 1 = 4 – 2 + 1 = 3&&
Answers: &&p(0) = 1, p(1) = 1, p(2) = 3&&
(ii) For polynomial &&p(t) = 2 + t + 2t^2 – t^3&&
- For &&p(0)&&: Substitute &&t=0&&.
&&p(0) = 2 + (0) + 2(0)^2 – (0)^3 = 2 + 0 + 0 – 0 = 2&& - For &&p(1)&&: Substitute &&t=1&&.
&&p(1) = 2 + (1) + 2(1)^2 – (1)^3 = 2 + 1 + 2 – 1 = 4&& - For &&p(2)&&: Substitute &&t=2&&.
&&p(2) = 2 + (2) + 2(2)^2 – (2)^3 = 2 + 2 + 2(4) – 8 = 4 + 8 – 8 = 4&&
Answers: &&p(0) = 2, p(1) = 4, p(2) = 4&&
(iii) For polynomial &&p(x) = x^3&&
- For &&p(0)&&: Substitute &&x=0&&.
&&p(0) = (0)^3 = 0&& - For &&p(1)&&: Substitute &&x=1&&.
&&p(1) = (1)^3 = 1&& - For &&p(2)&&: Substitute &&x=2&&.
&&p(2) = (2)^3 = 8&&
Answers: &&p(0) = 0, p(1) = 1, p(2) = 8&&
(iv) For polynomial &&p(x) = (x – 1)(x + 1)&&
- For &&p(0)&&: Substitute &&x=0&&.
&&p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1&& - For &&p(1)&&: Substitute &&x=1&&.
&&p(1) = (1 – 1)(1 + 1) = (0)(2) = 0&& - For &&p(2)&&: Substitute &&x=2&&.
&&p(2) = (2 – 1)(2 + 1) = (1)(3) = 3&&
Answers: &&p(0) = -1, p(1) = 0, p(2) = 3&&
Common Mistakes to Avoid 🚫
- Sign Errors: Be very careful with negative signs, especially when subtracting terms or raising negative numbers to a power.
- Order of Operations: Always calculate exponents first before you multiply, add, or subtract. For &&2t^2&&, you must square &&t&& before multiplying by 2.
Real-World Application 🌍
This skill is essential in science and engineering. For instance, the trajectory of a rocket can be modeled by a polynomial equation. By substituting a value for time (like &&t=10&& seconds), engineers can calculate the rocket’s exact height and velocity at that moment.
Frequently Asked Questions (FAQ)
Q: What does p(0) mean for a polynomial p(x)?
A: &&p(0)&& means you need to find the value of the polynomial &&p(x)&& when the variable &&x&& is equal to &&0&&. You do this by substituting &&0&& for every &&x&& in the expression.
Q: Can the value of a polynomial be zero?
A: Yes. When the value of a polynomial is zero for a specific number, that number is called a ‘zero of the polynomial‘. In our example, for &&p(x) = (x – 1)(x + 1)&&, the value &&p(1) = 0&&, so &&1&& is a zero of this polynomial.
Q: Is it necessary to expand p(x) = (x – 1)(x + 1) before substituting values?
A: No, it’s not necessary and often easier not to. You can directly substitute the value into the factored form. For example, for &&p(0)&&, you can calculate &&(0 – 1)(0 + 1) = (-1)(1) = -1&&, which is very fast.
Q: What is the order of operations for simplifying these expressions?
A: Always follow the order of PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), and finally Addition and Subtraction (from left to right).
Further Reading
- For more practice, refer to the official NCERT website for textbooks.
- To understand all concepts in this chapter, check out our summary of NCERT Solutions for Class 9 Maths Chapter 2 Polynomials.