jmc

One can find this by multiplying the two polynomials and computing the coefficients. Alternatively, we notice that the desired expression is simply the value of $a{x}^3+b{x}^2+cx+d$ evaluated at the point $x=2$: $a(2{)}^3+b(2{)}^2+c(2)+d=8a+4b+2c+d$. Since $(3x^2-5x+4)(7-2x)$ and $a{x}^3+b{x}^2+cx+d$ are equivalent expressions, they are equal for all real values of $x$. In particular, they are equal when $x=2$. So $8a+4b+2c+d=(3\cdot (2{)}^2-5\cdot (2)+4)(7-2\cdot (2))=6\cdot 3=\boxed{18}.$