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Mathematics · 2025 · 3 marks
CBSE 2025 · Region 6 · Set 1 · Q27
Find the zeroes of the polynomial p (x)=3 x^2-4 x-4. Hence, write a polynomial whose each of the zeroes is 2 more than zeroes of p (x).
Find the zeroes of the polynomial $\displaystyle \mathrm{p}(x)=3 x^{2}-4 x-4$. Hence, write a polynomial whose each of the zeroes is $\displaystyle 2$ more than zeroes of $\displaystyle \mathrm{p}(x)$.
Marking-scheme solution
\[p(x)=3 x^{2}-4 x-4
\]
Zeroes are \(\displaystyle -\frac{2}{3}\) and $\displaystyle 2$
New zeroes are \(\displaystyle \frac{4}{3}\) and $\displaystyle 4$
Sum of new zeroes \(\displaystyle =\frac{4}{3}+4=\frac{16}{3}\)
Product of new zeroes \(\displaystyle =\frac{4}{3} \times 4=\frac{16}{3}\)
Required polynomial is \(\displaystyle x^{2}-\frac{16 x}{3}+\frac{16}{3}\) or \(\displaystyle 3 x^{2}-16 x+16\)
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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.