Say we want to find the Taylor polynomial of order $2$ for $\sqrt{2+x}$ around $x = 10$.
Simply plugging in the numbers for a Taylor sum $\sum \dfrac{f^k(10)(x - 10)^k}{k!} $ with $n = 2$ then gives us a polynomial approximation of $\sqrt{2+x}$. Is it possible to give a different polynomial expansion than the one from the formula of $\sqrt{2+x}$ of order $2$ around $x = 10$? I personally don't understand how a different polynomial expression with the same order around the same $x$ is to be found. However, the problem I've found asks this precise question.
(Sorry for poor formatting)
The Taylor polynomial of order $2$ of the function $f(x):=\sqrt{2+x}$ is the unique polynomial $j_2$ of degree $\leq 2$ satisfying $$f(10+t)-j_2(t)= o(t^2)\qquad (t\to0)\ .$$ It gives a good approximation to the function value $f(x)$ when $x=10+t$ is near $10$, and the quality of this approximation increases tremendously as $t\to0$.
But, depending on your purpose, there are other polynomials of degree $\leq2$ approximating $f$ very well in the neighborhood of $x=10$, e.g. the polynomial $$h(x):=1.71446 + 0.205225 x - 0.00302609 x^2\ .$$ The following figure shows that that the maximal errror of $h$ in the $x$-interval $[8,12]$ is much smaller than the error of $j_2$.