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 $c=4$.