Polynomial knowing only the leading coefficient.

Note that $3=1^2+2$ , $11=3^2+2$ and $27=5^2+2$ . This means we have $f(x)=x^2+2+(x-k)(x-1)(x-3)(x-5)$ for some $k$ . Substituting in, we find

$f(-2)=216+105k$

$7f(6)=896-105k$

and so $f(-2)+7f(6)=\boxed{1112}$ .

Notice $f(x)=x^2+2$ for $x\in\{1,3,5\}$ . This means $(x-1)(x-2)(x-3)|f(x)-(x^2+2)$ , and because $f(x)$ is quartic with leading coefficient $1$ , $\frac{f(x)-(x^2+2)}{(x-1)(x-3)(x-5)}$ is linear with leading coefficient $1$ . In particular, we have that $\forall x,y \neq 1,3,5$ , $\frac{f(x)-(x^2+2)}{(x-1)(x-3)(x-5)}-\frac{f(y)-(y^2+2)}{(y-1)(y-3)(y-5)}=x-y$ . Taking $x=6,y=-2$ yields the solution: $7f(6)+f(-2)=1112$

If we set $f(x)=x^4+ax^3+bx^2+cx+d$ , we get the equations

$f(1)=1+a+b+c+d$

$f(3)=81+27a+9b+3c+d$

$f(5)=625+125a+25b+5c+d$

and we want to know

$f(-2)+7f(6)=9088+1504a+256b+40c+8d$

We will not be able to find a, b, c, d, because we only have three equations and 4 unknowns, but we are hoping that some linear combination $c_1f(1)+ c_3f(3) + c_5f(5)$ will be equal to $f(-2)+7f(6)$ (possibly apart from a constant).

So we want to solve

$1c_1 + 27c_3 + 125c_5 = 1504$ (a)

$1c_1 + 9c_3 + 25c_5 = 256$ (b)

$1c_1 + 3c_3 + 5c_5 = 40$ (c)

$1c_1 + 1c_3 + 1c_5 = 8$ (d)

These are 4 equations for 3 unknowns, but let's see if it works out. I used the last three eqations to find $c_1=7, c_3=-14, c_5=15$ and checked the first equation afterwards: $c_1 + 27c_3 + 125c_5 = 1504$ .

For the units we have $c_1 + 81c_3 + 625c_5 = 8248$ , which is 840 less than the 9088 required, so we need to add 840. We now know that $f(-2)+7f(6) = c_1f(1) +c_3f(3) + c_5f(5) + 840 = 7×3 -14×11 + 15 ×27 +840 = \boxed{1112}$ .