NCERT Solutions for Class 9 Maths Exercise 2.2 Question 2
Understanding the Question 🧐
This question asks us to evaluate four different polynomials at three specific values: 0, 1, and 2. The notation &&p(0)&&, &&p(1)&&, and &&p(2)&& is a shorthand way of asking, “What is the value of the polynomial when the variable is 0, 1, and 2?” We just need to substitute these values into each expression and calculate the result.
Find &&p(0)&&, &&p(1)&& and &&p(2)&& for each of the following polynomials:
(i) &&p(y) = y^2 – y + 1&&
(ii) &&p(t) = 2 + t + 2t^2 – t^3&&
(iii) &&p(x) = x^3&&
(iv) &&p(x) = (x – 1)(x + 1)&&
Part (i): &&p(y) = y^2 – y + 1&& 📝
- For &&p(0)&&: &&p(0) = (0)^2 – (0) + 1 = 0 – 0 + 1 = 1&&
- For &&p(1)&&: &&p(1) = (1)^2 – (1) + 1 = 1 – 1 + 1 = 1&&
- For &&p(2)&&: &&p(2) = (2)^2 – (2) + 1 = 4 – 2 + 1 = 3&&
Part (ii): &&p(t) = 2 + t + 2t^2 – t^3&& 📝
- For &&p(0)&&: &&p(0) = 2 + (0) + 2(0)^2 – (0)^3 = 2 + 0 + 0 – 0 = 2&&
- For &&p(1)&&: &&p(1) = 2 + (1) + 2(1)^2 – (1)^3 = 2 + 1 + 2(1) – 1 = 3 + 2 – 1 = 4&&
- For &&p(2)&&: &&p(2) = 2 + (2) + 2(2)^2 – (2)^3 = 2 + 2 + 2(4) – 8 = 4 + 8 – 8 = 4&&
Part (iii): &&p(x) = x^3&& 📝
- For &&p(0)&&: &&p(0) = (0)^3 = 0&&
- For &&p(1)&&: &&p(1) = (1)^3 = 1&&
- For &&p(2)&&: &&p(2) = (2)^3 = 8&&
Part (iv): &&p(x) = (x – 1)(x + 1)&& 📝
We can solve this by direct substitution.
- For &&p(0)&&: &&p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1&&
- For &&p(1)&&: &&p(1) = (1 – 1)(1 + 1) = (0)(2) = 0&&
- For &&p(2)&&: &&p(2) = (2 – 1)(2 + 1) = (1)(3) = 3&&
&&p(x) = (x – 1)(x + 1) = x^2 – 1&&.
Now, substituting the values is even easier:
&&p(0) = (0)^2 – 1 = -1&&
&&p(1) = (1)^2 – 1 = 0&&
&&p(2) = (2)^2 – 1 = 3&&
Conclusion and Summary ✅
Here is a summary of all the calculated values:
| Polynomial | &&p(0)&& | &&p(1)&& | &&p(2)&& |
|---|---|---|---|
| &&y^2 – y + 1&& | 1 | 1 | 3 |
| &&2 + t + 2t^2 – t^3&& | 2 | 4 | 4 |
| &&x^3&& | 0 | 1 | 8 |
| &&(x – 1)(x + 1)&& | -1 | 0 | 3 |
- To evaluate a polynomial at a certain value, replace every instance of the variable with that value.
- Always follow the correct order of operations (BODMAS/PEMDAS).
- Sometimes, simplifying the polynomial first can make the calculations quicker.
FAQ ❓
Q: What does it mean to find &&p(0)&&, &&p(1)&&, and &&p(2)&&?
A: It means you have to find the value of the polynomial ‘p’ for three different cases. First, you substitute the variable with 0 and calculate the result (this is &&p(0)&&). Then, you do the same for 1 (&&p(1)&&) and 2 (&&p(2)&&).
Q: What is a “zero of a polynomial”?
A: A “zero of a polynomial” is a value of the variable that makes the entire polynomial equal to zero. In this question, for the polynomial &&p(x) = (x – 1)(x + 1)&&, we found that &&p(1) = 0&&. This means that &&x=1&& is a zero of this polynomial.
Further Reading 📖
To learn more about evaluating polynomials and finding their zeros, you can refer to the official NCERT textbook for Class 9 Maths, Chapter 2. More resources are available on the NCERT website at https://ncert.nic.in/.
