116
Find the zeroes of the polynomial $p(x) = 3x^2 - 4x - 4$. Hence, write a polynomial whose each of the zeroes is 2 more than zeroes of $p(x)$.
Show SolutionHide Solution↓
$p(x) = 3x^2 - 4x - 4$.
Zeroes are $-\frac{2}{3}$ and 2 (1 mark).
New zeroes are $\frac{4}{3}$ and 4 ($\frac{1}{2}$ mark).
Sum of new zeroes = $\frac{4}{3} + 4 = \frac{16}{3}$ ($\frac{1}{2}$ mark).
Product of new zeroes = $\frac{4}{3} \times 4 = \frac{16}{3}$ ($\frac{1}{2}$ mark).
Required polynomial is $x^2 - \frac{16x}{3} + \frac{16}{3}$ or $3x^2 - 16x + 16$ ($\frac{1}{2}$ mark).
Zeroes are $-\frac{2}{3}$ and 2 (1 mark).
New zeroes are $\frac{4}{3}$ and 4 ($\frac{1}{2}$ mark).
Sum of new zeroes = $\frac{4}{3} + 4 = \frac{16}{3}$ ($\frac{1}{2}$ mark).
Product of new zeroes = $\frac{4}{3} \times 4 = \frac{16}{3}$ ($\frac{1}{2}$ mark).
Required polynomial is $x^2 - \frac{16x}{3} + \frac{16}{3}$ or $3x^2 - 16x + 16$ ($\frac{1}{2}$ mark).