Use middle term breaking.
Let, the factors be $(x+p)(x+q).$
So,we get $x^2+bx+c=(x+p)(x+q)=x^2+(p+q)x+pq.$
Comparing we get $b=(p+q)$ and $c=pq.$
So,you have to find numbers $p$ and $q$ such that $pq=c$ (constant term) and $p+q=b$.
If the question is of the form $ax^2+bx+c$ (like yours)
$pq≠c$ but $ac$.(Try to prove it,why?)
Now, coming back to your problem..
So,$c=-16\times 9=-144$ which is to be expressed in $pq$ form.Prime factors of $-144$ are $2,2,2,2,3,3$.Now this is to be arranged in the form such that $p+q=18$ and $pq=-144$.We see that $24$ and $-6$ are the required $p$ and $q$ $[24+(-6)=18]$ and $(24\times-6=-96).$
So,we can write,