NCERT Solutions for Class 9 Maths Exercise 2.3 Question 2
Understanding the Question 🧐
We need to find the remainder when one polynomial is divided by another. The easiest way to do this is by using the Remainder Theorem, which saves us from doing long division.
Find the remainder when &&x^3 – ax^2 + 6x – a&& is divided by &&x – a&&.
Applying the Remainder Theorem 📝
Step 1: Identify the polynomial and the divisor.
Let the polynomial be &&p(x) = x^3 – ax^2 + 6x – a&&.
The divisor is &&x – a&&.
Step 2: Find the zero of the divisor.
To find the zero, we set the divisor equal to 0 and solve for &&x&&:
&&x – a = 0 \implies x = a&&
Step 3: Substitute the zero of the divisor into the polynomial.
According to the Remainder Theorem, the remainder is &&p(a)&&. So, we substitute &&x = a&& into &&p(x)&&:
&&p(a) = (a)^3 – a(a)^2 + 6(a) – a&&
Step 4: Simplify the expression.
&&p(a) = a^3 – a(a^2) + 6a – a&&
&&p(a) = a^3 – a^3 + 6a – a&&
The &&a^3&& and &&-a^3&& terms cancel each other out.
&&p(a) = 6a – a = 5a&&
Conclusion: The remainder when &&x^3 – ax^2 + 6x – a&& is divided by &&x – a&& is &&5a&&.
FAQ ❓
Q: What is the zero of the divisor &&x – a&&?
A: To find the zero, you set the divisor equal to zero and solve for &&x&&. So, &&x – a = 0&& gives &&x = a&&. Therefore, the zero of the divisor is ‘&&a&&’.
Q: Can the remainder be an expression with a variable in it?
A: Yes. If the polynomial itself contains other variables or constants (like ‘&&a&&’ in this question), the remainder can also be an expression involving that variable. In this case, the remainder is &&5a&&.
Further Reading 📖
To learn more about the Remainder Theorem, 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/.
