f(u)=5u2+10u
\mathrm{f}(\mathrm{u})=5 \mathrm{u}^{2}+10 \mathrm{u}
\therefore \mathrm{f}(\mathrm{u})=0 \Rightarrow 5 \mathrm{u}=0 \text { or } \mathrm{u}+2=0
⇒u=0 or u=–2
\Rightarrow \mathrm{u}=0 \text { or } u=-2
\mathrm{f}(\mathrm{u})=5 \mathrm{u}^{2}+10 \mathrm{u}
It can be written as $5 \mathrm{u}(\mathrm{u}+2)$
∴f(u)=0⇒5u=0 or u+2=0\therefore \mathrm{f}(\mathrm{u})=0 \Rightarrow 5 \mathrm{u}=0 \text { or } \mathrm{u}+2=0
⇒u=0 or u=–2
\Rightarrow \mathrm{u}=0 \text { or } u=-2
So, the zeroes of $f(u)$ are $-2$ and 0 .
Sum of the zeroes $=-2+0=-2=\frac{-2 \times 5}{1 \times 5}=\frac{-10}{5}=\frac{-(\text { coefficient of } x)}{\left.\text { (coefficient of } u^{2}\right)}$
Product of zeroes $=-2 \times 0=0=\frac{0 \times 5}{1 \times 5}=\frac{-0}{5}=\frac{\text { constant term }}{\left(\text { coefficient of } u^{2}\right)}$