The number of polynomials having zeroes as -2 and 5 is

(d) Let p (x) = ax2 + bx + c be the required polynomial whose zeroes are -2 and 5.

$\therefore \quad$ Sum of zeroes $=\frac{-b}{a}$

$\Rightarrow$ $\frac{-b}{a}=-2+5=\frac{3}{1}=\frac{-(-3)}{1}$ ...(i)

and product of zeroes $=\frac{c}{a}$

$\Rightarrow \quad \frac{c}{a}=-2 \times 5=\frac{-10}{1}$ .....(ii)

From Eqs. (i) and (ii),

$a=1, b=-3$ and $c=-10$

$\therefore$ $p(x)=a x^{2}+b x+c=1 \cdot x^{2}-3 x-10$

$=x^{2}-3 x-10$

But we know that, if we multiply/divide any polynomial by any arbitrary constant. Then, the zeroes of polynomial never change.

$\therefore$ $p(x)=k x^{2}-3 k x-10 k$ [where, $k$ is a real number]

$\Rightarrow$ $p(x)=\frac{x^{2}}{k}-\frac{3}{k} x-\frac{10}{k}$ [where, $k$ is a non-zero real number]

Hence, the required number of polynomials are infinite i.e., more than 3.