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Polynomials
NCERT Exemplar Ch 02
CBSE_NCERT_EXEMPLAR_CH02
Grade 10

Question:

A quadratic polynomial, whose zeroes are $-3$ and $4$, is:

$x^2 - x + 12$
$x^2 + x + 12$
$\dfrac{x^2}{2} - \dfrac{x}{2} - 6$
$2x^2 + 2x - 24$

Step-by-Step Solution

Key Concept: Any quadratic polynomial with zeroes $\alpha, \beta$ is of the form $k[x^2 - (\alpha + \beta)x + \alpha \beta]$ for constant $k <br>eq 0$.
Stepwise Solution:

Sum of zeroes $\alpha + \beta = -3 + 4 = 1$. Product of zeroes $\alpha \beta = (-3)(4) = -12$. [0.5 Mark]

Polynomial $p(x) = k(x^2 - x - 12)$. Taking $k = 1/2$ gives $\dfrac{x^2}{2} - \dfrac{x}{2} - 6$. [0.5 Mark]

Marking Scheme:

• Finding sum and product of zeroes: 0.5 Mark
• Matching polynomial form with option: 0.5 Mark

Correct Answer: $\dfrac{x^2}{2} - \dfrac{x}{2} - 6$
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