#### Answer

$c=3$
$c=4$

#### Work Step by Step

If $y^2-4y+c=(y+p)(y+q)$, then
$c=p\bullet q$
$p+q=-4$
Since $c$ must be positive and the sum of $p$ and $q$ is a negative value, that means $p$ and $q$ must both be negative.
What two negative numbers add up to $-4$?
$-1$ and $-3$
$-2$ and $-2$
These are the possible values of $p$ and $q$.
Since $c=p\bullet q$, the product of the number combinations above will give us the values of $c$.
$-1\bullet-3=3$
$-2\bullet-2=4$
Therefore, $c=3$ and $c=4$.