NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 2: Polynomials - Exercise 2.1Question 2
Class 10 - Mathematics - CHAPTER 2: Polynomials - Exercise 2.1
Question. 2
A quadratic polynomial, whose zeroes are \(-3\) and \(4\), is
(A)
\(x^2 - x + 12\)
(B)
\(x^2 + x + 12\)
(C)
\(\dfrac{x^2}{2} - \dfrac{x}{2} - 6\)
(D)
\(2x^2 + 2x - 24\)
Detailed Answer with Explanation:
Step 1: Write the factor form of the polynomial.
If the zeroes are \(-3\) and \(4\), then the polynomial must be a multiple of
\[(x - (-3))(x - 4) = (x + 3)(x - 4)\]
Step 2: Expand the product.
\[(x + 3)(x - 4) = x^2 - 4x + 3x - 12\]
\[= x^2 - x - 12\]
Step 3: Compare with the given options.
The basic polynomial is \(x^2 - x - 12\).
Any non-zero constant multiple of this is also valid.
Step 4: Check the options.
- Option A: \(x^2 - x + 12\) → wrong (constant term differs).
- Option B: \(x^2 + x + 12\) → wrong (signs do not match).
- Option C: \(\dfrac{1}{2}(x^2 - x - 12) = \dfrac{x^2}{2} - \dfrac{x}{2} - 6\) → correct.
- Option D: \(2x^2 + 2x - 24\) → wrong, because it equals \(2(x^2 + x - 12)\), not a multiple of \(x^2 - x - 12\).
Conclusion: The correct polynomial is
\(\dfrac{x^2}{2} - \dfrac{x}{2} - 6\).