Why do I love Polynomials so much?

It is given that $f(x) = ax^4 + bx^2 + cx + d$ , where $a,b,c,d \in N$ . It is also given that:

$\begin{cases} f(1)+f(2) & = 17a+5b+3c+2d & = 44 &...(1) \\ f(2)+f(3) & = 97a + 13b + 5c + 2d & = 146 &...(2) \\ f(3) + f(4) & = 337a+25b+7c+2d & = 416 &...(3) \\ f(5) & = 625a+25b+5c+d & = 694 &...(4) \end{cases}$

From $(1):$ For $a,b,c,d \in N$ , we note that $a=1$ or $2$ . When $a=2$ , $\Rightarrow b=c=d=1$ . Substituting them in (2), we have $194 + 13 + 5 + 2 = \color{#D61F06}{214 \ne 146}$ , therefore $\color{#D61F06}{a \ne 2}$ but $\color{#3D99F6}{a = 1}$ . Substituting $a=1$ to the rest of the equations, we have:

$\begin{cases} (2): & 13b + 5c + 2d & = 49 &...(2a) \\ (3): & 25b+7c+2d & = 79 &...(3a) \\ (4): & 25b+5c+d & = 69 &...(4a) \end{cases}$

$\begin{array} {rlll} (3a)-(4a): & 2c+d = 10 \quad ...(5) & \Rightarrow \color{#3D99F6}{c < 5} \\ (2a): & 13b + c + 2(2c+d) = 49 & \Rightarrow 13b + c = 29 & \Rightarrow b = \begin{cases} 1 & \Rightarrow c = \color{#D61F06}{16 > 5 \text{ rejected}} \\ 2 & \Rightarrow c = \color{#3D99F6}{\space \space 3 < 5 \text{ accepted}}\end{cases} \\ (5): & 2(3)+d = 10 & \Rightarrow \color{#3D99F6}{d = 4} \end{array}$

Therefore,

$f(30)-f(20)+(a+b+c+d)abcd \\ = 810000 + 2(900)+3(30)+ 4 - [160000 + 2(400) + 3(20)+ 4] +(1+2+3+4)(1\times 2 \times 3 \times 4) \\ = 810000+1800+90+4-[160000+800+60+4] + 10(24) \\ = 811894 - 160864 + 240 = \boxed{651270}$