NCERT Solutions for Class 9 Maths Exercise 2.4 Question 3
Understanding the Question 🧐
In this question, we are working with the Factor Theorem again, but in a slightly different way. Instead of checking *if* something is a factor, we are told that &&x – 1&& *is* a factor. Our task is to use this piece of information to find the value of the unknown constant, &&k&&, in each polynomial.
The core idea is: Since we know &&x – 1&& is a factor of &&p(x)&&, the Factor Theorem guarantees that &&p(1) = 0&&. We will use this equation to solve for &&k&&. These ncert solutions will make the process clear.
Find the value of &&k&&, if &&x – 1&& is a factor of &&p(x)&& in each of the following cases.
Part (i): &&p(x) = x^2 + x + k&& 📝
Step 1: Apply the Factor Theorem.
Since &&x – 1&& is a factor, we know that &&p(1) = 0&&.
Step 2: Substitute &&x = 1&& into &&p(x)&& and set the expression to zero.
&&p(1) = (1)^2 + (1) + k = 0&&
Step 3: Solve the equation for &&k&&.
&&1 + 1 + k = 0&&
&&2 + k = 0&&
&&k = -2&&
Part (ii): &&p(x) = 2x^2 + kx + \sqrt{2}&& 📝
Step 1: Apply the Factor Theorem.
As &&x – 1&& is a factor, we must have &&p(1) = 0&&.
Step 2: Substitute &&x = 1&& into &&p(x)&& and set the expression to zero.
&&p(1) = 2(1)^2 + k(1) + \sqrt{2} = 0&&
Step 3: Solve the equation for &&k&&.
&&2(1) + k + \sqrt{2} = 0&&
&&2 + k + \sqrt{2} = 0&&
To isolate &&k&&, we move the other terms to the right side of the equation.
&&k = -2 – \sqrt{2}&&, which can also be written as &&k = -(2 + \sqrt{2})&&.
Part (iii): &&p(x) = kx^2 – \sqrt{2}x + 1&& 📝
Step 1: Apply the Factor Theorem.
Given that &&x – 1&& is a factor, it follows that &&p(1) = 0&&.
Step 2: Substitute &&x = 1&& into &&p(x)&& and set it to zero.
&&p(1) = k(1)^2 – \sqrt{2}(1) + 1 = 0&&
Step 3: Solve the equation for &&k&&.
&&k(1) – \sqrt{2} + 1 = 0&&
&&k – \sqrt{2} + 1 = 0&&
To find &&k&&, move the constant terms to the other side.
&&k = \sqrt{2} – 1&&
Part (iv): &&p(x) = kx^2 – 3x + k&& 📝
Step 1: Apply the Factor Theorem.
Since &&x – 1&& is a factor, we know that &&p(1) = 0&&.
Step 2: Substitute &&x = 1&& into &&p(x)&&.
&&p(1) = k(1)^2 – 3(1) + k = 0&&
Step 3: Solve the equation for &&k&&.
&&k – 3 + k = 0&&
Combine the terms with &&k&&.
&&2k – 3 = 0&&
&&2k = 3&&
&&k = \frac{3}{2}&&
Conclusion and Key Points ✅
By consistently applying the rule that &&p(1)=0&& since &&(x-1)&& is a factor, we found the value of &&k&& for each polynomial. The key is to transform the problem from polynomial theory into solving a simple linear equation.
- The problem statement gives you the main clue: &&x-1&& *is* a factor.
- This automatically means that &&p(1) = 0&&. This is the starting point for every part.
- Substitute &&x=1&& into the given polynomial.
- Set the resulting expression equal to zero.
- Use basic algebra to solve the equation for the unknown &&k&&.
FAQ
Q: Why do we set &&p(1) = 0&& in this question?
A: The problem states that &&(x-1)&& is a factor of the polynomial &&p(x)&&. According to the Factor Theorem, if &&(x-a)&& is a factor of &&p(x)&&, then &&p(a)&& must be equal to &&0&&. In this case, &&a=1&&, so we must have &&p(1) = 0&&.
Q: How is this question different from the previous ones in this exercise?
A: In the previous questions, we were asked to check IF a given polynomial was a factor. In this question, we are TOLD that it IS a factor, and we have to use that information to find the value of an unknown constant, &&k&&.
Q: What is the value of &&k&& if &&(x-1)&& is a factor of &&p(x) = x^2 + x + k&&?
A: The value of &&k&& is &&-2&&. We set &&p(1)=0&&, which gives the equation &&1 + 1 + k = 0&&. Solving for &&k&& gives &&k = -2&&.
Q: What if the factor was &&(x+1)&& instead?
A: If the factor was &&(x+1)&&, its zero would be &&x = -1&&. In that case, we would apply the Factor Theorem by setting &&p(-1) = 0&& and solving for &&k&&.
Q: How do you solve for &&k&& in the polynomial &&p(x) = kx^2 – 3x + k&&?
A: Since &&(x-1)&& is a factor, we set &&p(1)=0&&. This gives &&k(1)^2 – 3(1) + k = 0&&, which simplifies to &&k – 3 + k = 0&&, or &&2k – 3 = 0&&. Solving this equation gives &&k = \frac{3}{2}&&.
Further Reading
For a stronger foundation in polynomial factors, refer to your Class 9 Maths textbook. The official NCERT materials are the best source and can be found on their website: https://ncert.nic.in/.
