10 degrees above Zero

$f(x)=\color{#D61F06}{A}\prod_{r=1}^{10}(x-r)$

$\implies \dfrac{f(12)}{f(11)}=\dfrac{\cancel{\color{#D61F06}{A}}\displaystyle\prod_{r=1}^{10}(12-r)}{\cancel{\color{#D61F06}{A}}\displaystyle\prod_{r=1}^{10}(11-r)}=\dfrac{11\cancel{!}}{\cancel{10!}}=11$

$\implies f(12)=11\cdot f(11)=11\cdot 11=\boxed{121}$

Since all the roots of the polynomial are known, $f(x)$ can be written as $f(x)=k\prod_{i=1}^{10}(x-i)$ .

Evaluating $f(11)$ , we get the relation $f(11)=k\times 10! = 11$ , which implies $k=\frac{11}{10!}$ .

Simple substitution would now give $f(12)=k\prod_{i=1}^{10}(12-i)=\frac{11}{10!}11!=11^2=\boxed{121}$ .

$f(x)$ has roots $1,2,3,4,5,6,7,8,9,10$ . Therefore,

$f(x) = k(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10)$

$f(11)=11\\ k(11-1)(11-2)(11-3)(11-4)(11-5)(11-6)(11-7)(11-8)(11-9)(11-10)=11\\ k(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)=11\\ \implies k =\dfrac{11}{10!}$

We then calculate $f(12)$ :

$f(12)=\dfrac{11}{10!}(12-1)(12-2)(12-3)(12-4)(12-5)(12-6)(12-7)(12-8)(12-9)(12-10)\\ =\dfrac{11}{10!}(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)\\ =\dfrac{11^2}{\color{#D61F06}{\cancel{\color{#333333}{10!}}}}(\color{#D61F06}{\cancel{\color{#333333}{10!}}})\\ =\boxed{121}$

The real purpose of me writing this solution:

To tell you that

$f(x)=\dfrac{11}{3628800}x^{10}-\dfrac{121}{725760}x^9+\dfrac{121}{30240}x^8-\dfrac{1331}{24192}x^7+\dfrac{82643}{172800}x^6\\-\dfrac{94501}{34560}x^5+\dfrac{3758623}{362880}x^4-\dfrac{925045}{36288}x^3+\dfrac{1948463}{50400}x^2-\dfrac{81191}{2520}x+11$