When all's unknown

It is noted that the Taylor series expansion of $f(x)$ about $x=a$ is same as $f(x) = ax^3 + bx^2 + cx + d$ . (I actually found out by evaluating $\sum_{n=0}^3 \frac {f^{(n)}(a)(x-a)}{n!}$ ) . Therefore, we have:

$\frac{d}{6}(x- a)^3 + (3a + b + 2d)(x- a)^2 + (3a^2 + c + 6d)(x - a) = ax^3 + bx^2 + cx + d$

Equating coefficients, we have:

For $x^3$ : $\quad \dfrac d6 = a \implies \color{#3D99F6}{d = 6a} \implies \dfrac d6 (x-a)^3 = a(x-a)^3$

For $x^2$ : $\quad -3a^2 + 3a + b + 2d = b \implies - 3a^2 + 15a = 0 \implies \color{#3D99F6}{a = 5} \implies \color{#3D99F6}{d = 30}$

For $x$ :

$\begin{aligned} 3a^3 -2a(3a + b + 2d) + 3a^2 + c + 6d & = c \\ 3a^2 -2(15a + b) + 3a + 36 & = 0 \\ 75 - 150 - 2b + 15 + 36 & = 0 \\ \implies \color{#3D99F6}{b} & \color{#3D99F6}{=-12} \end{aligned}$

For constant:

$\begin{aligned} -a^4 + a^2(3a + b + 2d) - a(3a^2 + c + 6d) & = d \\ -a^3 + a(3a + b + 2d) - (3a^2 + c + 6d) & = 6 \\ -125 + 315 - 255 - c & = 6 \\ \implies \color{#3D99F6}{c} & \color{#3D99F6}{=-71} \end{aligned}$

$\implies abcd = (5)(-12)(-71)(30) = \boxed{127800}$

Recall the general form of the Taylor Series expansion of any function $f(x)$ at $x\: = \: a$ .

$\large { \sum_{n=0}^{\infty} \frac{f^{(n)} (a)}{n!}(x-a)^{n} }$

For polynomial functions of degree $p$ , this expansion will terminate when $n= \: p$ , as the $(p)^{th}$ derivative of this polynomial will be constant.

Let us first see the derivatives of $f(x)$ .

$f^{(0)} (x) = \: ax^3 \: + \: bx^2 \: +\: cx\: + \: d$

$f^{(1)} (x)= \: 3ax^2 \: + \: 2bx \: +\: c$

$f^{(2)} (x)= \: 6ax \: + \: 2b$

$f^{(3)} (x)= \: 6a$

$f^{(4)} (x)= \: 0$ ...

Substituting $x=\: a$ , we get

$f^{(0)} (a) = \: a^4 \: + \: a^2b \: +\: ac\: + \: d$

$f^{(1)} (a)= \: 3a^3 \: + \: 2ab \: +\: c$

$f^{(2)} (a)= \: 6a^2 \: + \: 2b$

$f^{(3)} (a)= \: 6a$

Now, plugging this into the summation, we get

$f(x) = f^{(0)} (a) \: +\: f^{(1)} (a) \cdot (x\: - \: a) \: + \: \frac{ f^{(2)} (a)}{2!} \cdot (x\: - \: a)^2 \: + \: \frac{ f^{(3)} (a)}{3!} \cdot (x\: - \: a)^3 \:$

$f(x) = {(a^4+a^2b+ac+d)} \: + \: {(3a^3+2ab+c)}\cdot(x - a) \: +\: {\frac{6a^2+2b}{2!}}\cdot (x - a)^2 \:+\: \frac{6a}{3!} \cdot (x - a)^3$

$f(x) = (a^4+a^2b+ac+d) \: + \: (3a^3+2ab+c)\cdot(x - a) \: +\: (3a^2 + b)(x - a)^2 \:+\: a \cdot (x - a)^3$

From here, we get the correspondence from the given expansion in the problem

(1) $a\:=\: \frac{d}{6}$

(2) $3a^2 + b\: =\: 3a + b + 2d$

(3) $3a^3+2ab+c \: =\: 3a^2 + c + 6d$

(4) $a^4+a^2b+ac+d \:=\: 0$

Modifying (1) a bit and using $d = 6a$ to substitute for (2)..

we get

$3a^2 \: =\: 15a$ , or $\boxed{ a \: = \: 5}$ . Then we get $\boxed { d \: =\: 30 }$ .

Substituting both for (3), we get $\boxed { b \: =\: -12 }$ . Ultimately, we will get $\boxed { c \: =\: -71 }$ by substituting all these values into (4).

And bingo, $\boxed {5 \cdot -12 \cdot -71 \cdot 30 \: = \: 127800 }$

That is, the polynomial $f(x)$ is actually

$\large { 5x^3 - 12x^2 - 71x + 30}$

and its Taylor series expansion about $x\: = \: 5$ is

$\large { 5(x-5)^3 + 63(x-5)^2 + 184(x-5) }$