Find the zeros of the quadratic polynomial

We have:

$f(x)=2 x^{2}-11 x+15$

$=2 x^{2}-(6 x+5 x)+15$

$=2 x^{2}-6 x-5 x+15$

$=2 x(x-3)-5(x-3)$

$=(2 x-5)(x-3)$

$\therefore f(x)=0=>(2 x-5)(x-3)=0$

$=>2 x-5=0$ or $x-3=0$

$=>x=\frac{5}{2}$ or $x=3$

So, the zeroes of $f(x)$ are $\frac{5}{2}$ and 3 .

Sum of the zeroes $=\frac{5}{2}+3=\frac{5+6}{2}=\frac{11}{2}=\frac{-(\text { coefficient of } x)}{\left(\text { coefficient } t x^{2}\right)}$

Product of the zeroes $=\frac{5}{2} \times 3=\frac{15}{2}=\frac{\text { constant term }}{\text { (coefficient of } x^{2} \text { ) }}$