A quadratic polynomial,

Solution:

(c) Let $a x^{2}+b x+c$ be a required polynomial whose zeroes are $-3$ and 4 .

Then, sum of zeroes $=-3+4=1$ $\left[\because\right.$ sum of zeroes $\left.=\frac{-b}{a}\right]$

$\Rightarrow$ $\frac{-b}{a}=\frac{1}{1} \Rightarrow \frac{-b}{a}=-\frac{(-1)}{1}$ ... (i)

and product of zeroes $=-3 \times 4=-12$ $\left[\because\right.$ product of zeroes $\left.=\frac{c}{a}\right]$

$\Rightarrow$ $\frac{c}{a}=\frac{-12}{1}$ ...(ii)

From Eqs. (i) and (ii), 

$a=1, b=-1$ and $c=-12$

$=a x^{2}+b x+c$

$\therefore$ Required polynomial $=1 \cdot x^{2}-1 \cdot x-12$

$=x^{2}-x-12$

$=\frac{x^{2}}{2}-\frac{x}{2}-6$

We know that, if we multiply/divide any polynomial by any constant, then the zeroes of polynomial do not change.

Alternate Method

Let the zeroes of a quadratic polynomial are α = – 3 and β = 4.

Then, sum of zeroes =α + β = -3+4=1 and product of zeroes = αβ = (-3) (4) = -12