$y=x^2+bx+256$ cuts the $x-\text{axis}$ at $(h,0)$  and $(k,0)$. If $h$ and $k$ are integers then what is the least value of  $b$ is

$h$ and $k$ are the roots of  $x^2+bx+256=0$

$h+k=-b$ and  $h\times k=256$

$b=-(h+k)$ and  $h\times k=256$

Maximum value of $h+k$ is minimum value of $b$.

$\begin{array}{l}h\times k=256=1\times 256\\ =2\times 128\\ =4\times 64\\ =8\times 32\\ =16\times 16\end{array}$

$h+k=1+256=257$  is the maximum value

$\therefore$ least value of  $b$ is $-257$