Base B Number System Factor

$297_B = 2B^2 + 9B + 7$. $792_B = 7B^2 + 9B + 2$. Since $297_B$ is a factor of $792_B$, $7B^2 + 9B + 2 = k(2B^2 + 9B + 7)$. For $k = 2$: $7B^2 + 9B + 2 = 4B^2 + 18B + 14 \implies 3B^2 - 9B - 12 = 0 \implies B^2 - 3B - 4 = 0 \implies B = 4$ (Invalid, base must be > 9). For $k = 3$ (or rearranging $5B^2 - 5 = k(2B^2 + 9B + 7)$): $5B^2 - 5 = 2(2B^2 + 9B + 7) \implies B^2 - 18B - 19 = 0 \implies (B - 19)(B + 1) = 0 \implies B = 19$.