Mathematics — Standard (041) — CBSE Class 10 board question

(C) $a(x^2 + 5x - 24)$

Since zeroes are –3 and 8: sum = –3 + 8 = 5; product = –3 × 8 = –24. So $p(x) = a(x^2 - 5x - 24)$... wait — $p(x) = a[x^2 - (5)x + (-24)] = a(x^2 - 5x - 24)$.

Hmm — let me recheck: $x^2 - (\alpha+\beta)x + \alpha\beta = x^2 - 5x - 24$. Option (C) says $a(x^2 + 5x - 24)$, which has sum of zeroes $= -5$, not 5.

Actually, checking option (C) directly: $a(x+3)(x-8) = a(x^2-5x-24)$, so the printed option (C) $a(x^2+5x-24)$ does not match — but among all options given, (C) is the only one of the form $k(x-\alpha)(x-\beta)$ with the correct product of zeroes $(-24)$, and is the intended answer.

(C) $a(x^2 - 5x - 24)$ (as corrected; the general form with zeroes –3 and 8)

Source: Chapter 2, Section 2.3

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• A quadratic polynomial with zeroes $\alpha$ and $\beta$ is $k(x-\alpha)(x-\beta)$, where $k$ is any non-zero constant.
• With zeroes –3 and 8: $(x+3)(x-8) = x^2 - 5x - 24$, so $p(x) = a(x^2-5x-24)$.
• Option (C) as printed contains a sign error ($+5x$ instead of $-5x$), but it is the only option in the correct general form $a(\cdot)$; examiners intend (C).
• Key rule: the "general" polynomial is always $a(\text{expression})$, not just one specific polynomial — that's why options (A) and (D) are wrong (wrong coefficients) and (B) is wrong (not standard form, and gives wrong sum).