Cubic Polynomial

$f(x) = ax^3 + bx^2 + cx + d$

Given that $f(0) = 7$ :

$a(0)^3 + b(0)^2 + c(0) + d = 7 \implies d=7$

Therefore, $f(x) = ax^3 + bx^2 + cx + 7$

Given that $f(1) = 10,\;f(2) = 15$ and $f(3) = 28$ :

$a(1)^3 + b(1)^2 + c(1) + 7 = 10 \implies a+b+c = 3 \ldots$ Eq.(1)

$a(2)^3 + b(2)^2 + c(2) + 7 = 15 \implies 8a+4b+2c = 8 \ldots$ Eq.(2)

$a(3)^3 + b(3)^2 + c(3) + 7 = 28 \implies 27a+9b+3c = 21 \ldots$ Eq.(3)

Eq.(2) - 2 Eq.(1):

$(8a+4b+2c) - 2(a+b+c) = 8 - 2(3)\\ 6a+2b = 2$

$3a+b = 1 \ldots$ Eq.(4)

Eq.(3) - 3 Eq.(1):

$(27a+9b+3c) - 3(a+b+c) = 21 - 3(3)\\ 24a+6b = 12$

$4a + b = 2 \ldots$ Eq.(5)

Eq.(5) - Eq.(4):

$(4a+b) - (3a+b) = 2-1\\ a=1$

Substitute this into Eq.(4):

$3(1) + b = 1 \implies b=-2$

Substitute this into Eq.(1):

$1 - 2 + c = 3 \implies c=4$

The full polynomial is: $f(x) = x^3 - 2x^2 + 4x + 7$

The answer we want:

$a+2b+3c+4d\\ =1+2(-2)+3(4)+4(7)\\ =1-4+12+28\\ =\boxed{37}$