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Mathematics · 2025 · 3 marks
CBSE 2025 · Region 5 · Set 3 · Q26
Find the zeroes of the polynomial p (x)=6 x^2-5 x-1. Hence, obtain a polynomial each of whose zeroes is three times the zeroes of p (x).
Find the zeroes of the polynomial $\displaystyle \mathrm{p}(x)=6 x^{2}-5 x-1$. Hence, obtain a polynomial each of whose zeroes is three times the zeroes of $\displaystyle \mathrm{p}(x)$.
Marking-scheme solution
\[\begin{aligned}
p(x) & =6 x^{2}-5 x-1 \\
& =(x-1)(6 x+1)
\end{aligned}
\]
∴ Zeroes are $\displaystyle 1$, \(\displaystyle -\frac{1}{6}\)
New zeroes are \(\displaystyle 3,-\frac{1}{2}\)
Sum of new zeroes \(\displaystyle =3+\left(-\frac{1}{2}\right)=\frac{5}{2}\)
Product of new zeroes \(\displaystyle =3 \times\left(-\frac{1}{2}\right)=-\frac{3}{2}\)
∴ Required polynomial is \(\displaystyle x^{2}-\frac{5}{2} x-\frac{3}{2}\) or \(\displaystyle 2 x^{2}-5 x-3\)
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CBSE Class 10 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.