Answer
5, 4
Work Step by Step
$3x^2-8x+c$
The polynomial above can be re-written as $(3x+a)(x+b)$
$(3x+a)(x+b)$
$3x*x+3x*b+a*x+a*b$
$3x^2+3bx+ax+ab$
$3x^2=3x^2$
$3bx+ax=-8x$
$ab=c$
$3bx+ax=-8x$
$3b+a=-8$
Since we want positive values of $c$, then $a$ and $b$ have the same sign.
$b=-3$
$3b+a=-8$
$3*-3+a=-8$
$-9+a=-8$
$a=1$
$b=-2$
$3b+a=-8$
$3*-2+a=-8$
$-6+a=-8$
$a=-2$
$b=-1$
$3b+a=-8$
$3*-1+a=-8$
$-3+a=-8$
$a=-5$
$b=0$
$3b+a=-8$
$3*0+a=-8$
$a=-8$
$b=1$
$3b+a=-8$
$3*1+a=-8$
$3+a=-8$
$a=-11$
$b\ne-3$ as the sign of $a$ is positive. Also, for this same reason, $b\ne0$ and $b\ne1$. Thus, the only values of $b$ are -2 and -1.
The respective values of $a$ (for $b=-2$ and $b=-1$) are $-2$ and $-5$.
With $a=-2$ and $b=-2$, $c=-2*-2$, or $4$.
With $a=-5$ and $b=-1$, $c=-5*-1$, or $5$.
Thus, the only possible positive values of $c$ are $5$ and $4$.
