In the Theorem below put $\rm\:p\!=\!2,\ c=-5,\ d = 1\:$ to deduce that $\rm\:mod\ 16,\ (x\!+\!5)(x\!-\!1)\:$ has roots $\rm\:x \equiv -5,1\pmod 8,\:$ which are $\rm\:x,x\!+\!8 \equiv -5,1,3,9\pmod{16}.$ It has $\rm\,4\ (\!vs.\,2)$ roots by $\rm\:x\!+\!5 \equiv x\!-\!1\pmod{2},\:$ so both are divisible by $2$, so the other need be divisible only by $\rm8\ (\!vs. 16).$ Choosing larger primes $\rm\:p\:$ yields a quadratic with as many roots as you desire. These matters are much clearer $\rm\:p$-adically, e.g. google Hensel's Lemma.
Theorem $\ $ If prime $\rm\:p\:|\:c\!-\!d\:$ but $\rm\:p^2\nmid c\!-\!d\:$ then $\rm\:(x\!-\!c)(x\!-\!d)\:$ has $\rm\:2\!\;p\:$ roots mod $\rm\:p^4,\:$ namely $\rm\:x \equiv c+j\,p^3\:$ and $\rm\: x\equiv d+j\,p^3,\:$ for $\rm\:0\le j \le p\!-\!1.$
Proof $\ $ Note $\rm\: a = x\!-\!d,\ b = x\!-\!c\:$ satisfy the hypotheses of the Lemma below, thus we deduce $\rm\:p^4\:|\:(x\!-\!c)(x\!-\!d)\iff p^3\:|\:x\!-\!c\:$ or $\rm\:p^3\:|\:x\!-\!d,\:$ i.e. $\rm\:x\equiv c,d\pmod{p^3}.\:$ This yields the claimed roots $\rm\:mod\,\ p^4,\:$ which are all distinct since $\rm\:c+jp^3\equiv d+kp^3\:$ $\Rightarrow$ $\rm\:p^4\:|\:c\!-\!d+(j\!-\!k)p^3\:$ $\Rightarrow$ $\rm\: p^3\:|\:c\!-\!d,\:$ contra hypothesis. $\quad$ QED
Lemma $\ $ If prime $\rm\:p\:|\:a\!-\!b\:$ but $\rm\:p^2\nmid a\!-\!b\:$ then $\rm\:p^4\:|\:ab\iff p^3\:|\:a\ $ or $\rm\:p^3\:|\:b.$
Proof $\rm\,\ (\Rightarrow) \ \ p\:|\:ab\:\Rightarrow\:p\:|\:a\:$ or $\rm\:p\:|\:b,\:$ so $\rm\:p\:|\:a,b\:$ by $\rm\:a\equiv b\pmod p.$ But not $\rm\:p^2\:|\:a,b\:$ else $\rm\:p^2\:|\:a\!-\!b.\:$ So one of $\rm\:a,b\:$ is not divisible by $\rm\:p^2,\:$ hence the other is divisible by $\rm\:p^3.$
$(\Leftarrow)\ \ $ As above, $\rm\:p\:|\:a,b.\:$ Since one is divisible by $\rm\:p^3,\:$ then $\rm\:p^4\:$ divides their product. $\ \ $ QED