polynomial

$\left(x+2\right)\left(3x^2+x+5\right)$

$\mathrm{Distribute\:parentheses}$

$x\cdot \:3x^2+xx+x\cdot \:5+2\cdot \:3x^2+2x+2\cdot \:5$

$3x^3+7x^2+7x+10$

Sub these in $A + B + 2C + 5D$

$3+3+2(7)+5(10)$

$=70$

-Vinculum

$$\displaystyle 3x^3 + x^2 + 5x + 6x^2 + 2x + 10, which \ equals \ 3x^3 + 7x^2 + 7x + 10. The \ answer \ is \ 3 + 1 + 14 + 50, which \ is \ 68.$$

I think both of you guys messed up :)