The Constant Term

Since the roots form an arithmetic progression, there are numbers $a$ and $d$ such that the roots are $a-d$ , $a$ and $a+d$ . By one of Vieta's formulas, we see that $-3 = -(a-d) -a - (a+d)$ , implying $a = 1$ . By another one of Vieta's formulas, we see that

$-13 = (a-d)(a) + (a-d)(a+d) + (a)(a+d),$

and substituting $1$ for $a$ gives us

$-13 = (1-d) + (1-d)(1+d) + (1+d).$

This equation simplifies to $-13 = 3 - d^2$ , so $d^2 = 16$ and $d = \pm 4$ .

Whether $d$ is $4$ or $-4$ , the three roots are $-3$ , $1$ , and $5$ . Finally, by another of Vieta's formulas, the product of the three roots must be $-c$ , and so $c = -(-3)(1)(5) =$ 15.