Most valuable coefficient

We can use the Method of Differences .

The difference table for the polynomial is given above.
$D_{1}(x) = f(x+1) - f(x)$
$D_{2}(x) = D_{1}(x+1) - D_{1}(x)$
$D_{n}(x) = D_{n-1}(x+1) - D_{n-1}(x)$

We know that for a polynomial of degree n, and leading co-efficient a, $D_{n}(x)$ is a constant and is equal to $n! \times a$ .
$\therefore D_{5}(x) = 5! \times 3 = 120 \times 3 = 360$
Using this table $f(20)$ can be computed.

$\therefore f(20) =360 + 50 + 20 + 25 + 90 + 365 = \boxed{910}$

We can use Lagrangian interpolation . There is a unique polynomial of degree $n-1$ passing through $n$ specific points (there are $n$ conditions to be satisfied by the $n$ coefficients). Lagrangian interpolation gives us a quick method of determining that polynomial.

Define $\begin{array}{rcl} p_{15}(x) & = & (x-16)(x-17)(x-18)(x-19) \\ p_{16}(x) & = & (x-15)(x-17)(x-18)(x-19) \\ p_{17}(x) & = & (x-15)(x-16)(x-18)(x-19) \\ p_{18}(x) & = & (x-15)(x-16)(x-17)(x-19) \\ p_{19}(x) & = & (x-15)(x-16)(x-17)(x-18) \end{array}$ The whole point of these quartic polynomials is that, for example, $p_{16}(x)$ vanishes at $15,17,18,19$ , but is nonzero at $16$ . We can use this property to define $\begin{array}{rcl} g(x) & = & \dfrac{125}{p_{15}(15)}p_{15}(x) + \dfrac{150}{p_{16}(16)}p_{16}(x) + \dfrac{210}{p_{17}(17)}p_{17}(x) + \dfrac{275}{p_{18}(18)}p_{18}(x) + \dfrac{365}{p_{19}(19)}p_{19}(x) \\ & = & \dfrac{125}{24}p_{15}(x) + \dfrac{150}{-6}p_{16}(x) + \dfrac{210}{4}p_{17}(x) + \dfrac{275}{-6}p_{18}(x) + \dfrac{365}{24}p_{19}(x) \end{array}$ Then $g(x)$ is the unique polynomial of degree $4$ taking the values $125,150, 210, 275, 365$ at the points $15,16,17,18,19$ , and so the quintic polynomial we want is $f(x) \; = \; 3(x - 15)(x-16)(x-17)(x-18)(x-19) + g(x) \;.$ Thus $\begin{array}{rcl} f(20) & = & 3 \times120 + g(20) \\ & = & 360 + \dfrac{125}{24} p_{15}(20) - 25p_{16}(20) + \dfrac{105}{2}p_{17}(20) - \dfrac{275}{6} p_{18}(20) + \dfrac{365}{24}p_{19}(20) \\ & = & 360 + \dfrac{125}{24} \times 24 - 25 \times 30 + \dfrac{105}{2}\times40 - \dfrac{275}{6} \times60 + \dfrac{365}{24}\times120 \\ & = & 910 \end{array}$ The merit of this method is that it does not require the values $15,16,17,18,19$ to be consecutive; the Method of Differences does.