Answer
positive; positive
Work Step by Step
Given:
$x^2+bx+c=(x+p)(x+q)$
Note that when multiplied using FOIL method,
$(x+p)(x+q)
\\= x(x)+xq+px+pq
\\=x^2+(p+q)x=pq$
Thus,
$x^2+bx+c=(x+p)(x+q) = x^2+(p+q)x+pq$
This means that
$b=p+q
\\c=pq$
In $c=pq$, if $c$ is positive, then either both $p$ and $q$ are positive or both are negative (since the product of two numbers is positive only when the two numbers have the same sign).
However, since $b$ is also positive, then $p$ and $q$ cannot be negative; otherwise, their sum, $p+q$, which is known to be equal to $b$, will also be negative.
Therefore, if both $b$ and $c$ are positive, then $p$ and $q$ must also be both positive.
