SOLUTION —
Let $k$ be the number to be added to the given polynomial. Then the polynomial becomes $x^2-5 x+(4+k)$.
As 3 is the zero of the polynomial, we get :
$\begin{aligned}& & (3)^2-5(3)+(4+k) & =0 \\\Rightarrow & & (4+k) & =15-9 \\\Rightarrow & & 4+k & =6 \\\Rightarrow & & k & =2\end{aligned}$
Thus, 2 is to be added to the polynomial.