Try To Solve This Problem Without Expanding

Because $211 = 3^5 - 2^5, 65 = 3^4-2^4, 19 = 3^3- 2^3, 5 = 3^2-2^2,e = 3^1-2^1$

We consider finding values of $f(2)$ and $f(3)$ for $f(x) = (2x^3 - 3x^2 + 6x + 11)(5x^2 - 9x + 1)$

$f(2) = \left ( (2(3^3) - 3(3^2) + 6(3) + 11)(5(3^2)-9(3)+1)\right ) = 81$

$f(2) = \left ( (2(3^3) - 3(3^2) + 6(3) + 11)(5(3^2)-9(3)+1)\right ) = 1064$

So the expression equals to $f(3) - f(2) + f = 1064 - 81 + 11 = \boxed{994}$

Note that $211=3^5-2^5$ , $65=3^4-2^4$ , $19=3^3-2^3$ , $5=3^2-2^2$ , and $1=3-2$ . Thus, if $f(x)$ is the function given, the value we are looking for is $f(3)-f(2)+f$ . We can calculate that $f(3)=1064$ , $f(2)=81$ , and $f$ is the constant term, which is $11\cdot 1=11$ . Then, the answer is $1064-81+11=\boxed{994}$ .

Just expand the expression and you get the expression as $10x^{5}-33x^{4}+59x^{3}-2x^{2}-93x+11$

Comparing this with $ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f$ , we get the values as $a=10, b=(-33), c=59, d=(-2), e=(-93) and f=11$

$211a+65b+19c+5d+e+f$ $= (211\times 10) + (65\times (-33))+(19\times 59)+(5\times (-2))-93+11$ $= 2110-2145+1121-10-93+11 = \boxed{994}$

Doing the expansion $(2x^3 -3x^2+6x+11)(5x^2 - 9x+1)=10x^5-33x^4+59x^3-2x^2-93x+11,$ so $a=10\\b=-33\\c=59\\d=-2\\e=-93\\f=11.$ Thus $211a+65b+19c+5d+e+f=\\211(10)+65(-33)+19(59)+5(-2)+(-93)+11=\\\boxed{994}$

1. To get the term with $x^5$ as variables, we can only multiply $2x^3$ and $5x^2$ . Hence 10 is the coefficient of $x^5$ . Or shortly we say $a=10$
2. We can make a multiplication match to get the term with $x^4$ , i.e: $2x^3$ and $-9x$ , and $-3x^2$ and $5x^2$ are two matches I mean. Adding the coefficient of both matches gives us the coefficient of $x^4$ , $b=-33$ .
3. $x^3$ has varier possibilities of matches: $2x^3$ with $1$ , $-3x^2$ with $-9x$ , and $6x$ with $5x^2$ . Analogous to point (1) and (2), we get $c=59$
4. Matching $-3x^2$ with $1$ , $6x$ with $-9x$ , and $11$ with $x^2$ gives us $d=-2$
5. Since we've understand the logic, we can say directly $e=-93$
6. The last, $f=11$
7. Simply, $211\times10+65\times(-33)+19\times59+5\times(-2)+(-93)+11=2110-2145+1121-10-93+11=\boxed {994}$

Denote $g(x)=(2x^3-3x^2+6x+11)(5x^2-9x+1)=ax^5+bx^4+cx^3+dx^2+ex+f$ and let $\alpha=211a+65b+19c+5d+e+f$ Then $g(3)-\alpha=32a+16b+8c+4d+2e=g(2)-f=g(2)-g(0)$ So $\alpha=g(3)-g(2)+g(0)=1064-81+11=994$

Its just simple and straight. See how many terms will have x^4 and add the coefficients to get b and so on and finally calculate the required quantity.

I don't really know how to write long division in my solution. If anyone could help me, great. Well, I performed long division with the polynomial $10x^{5} + bx^{4} + cx^{3} + dx^{2} + ex + f$ . I divided it by $2x^{3} - 3x^{2} + 6x + 11$ , putting first as the result the polynomial $5x^{2} - 9x + 1$ . By this way, I found $b$ , $c$ and $d$ ( $a$ , $e$ and $f$ are very clear). I was able to form the following equations:

$(15 + b)x^{4} + 18x^{4} = 0 \rightarrow b = -33$ .

$(c - 57)x^{3} - 2x^{3} = 0 \rightarrow c = 59$ .

$(d + 2)x^{2} + 0x^{2} = 0 \rightarrow d = -2$ .

Clearly, and without necessity of expansion, we can see that $a = 10$ , $e = -93$ and $f = 11$ . So substituting values on the expression gives the answer $\boxed {994}$ .

(2x^3-3x^2+6x+11)(5x^2-9x+11)=10x^5-33x^4+59x^3-2x^2-93x+11 =>a=10;b=-33;c=59;d=-2;e=-93;f=11 =>211a+65b+19c+5d+e+f=994

Multiply the two terms and compare the coefficients with the given equation. Then you get a=10, b=-33, c= 59 d= -2 e= -93 and f = 11 Substitute these values in the given equation you get the result 994

994

by multiplying the given expression then cmparing coefficients ,we have a=10 b= -33 c=59 d= -2 e= -93 f=11 putting these values in given expression ,we have 211a+65b+19c+5d+e+f=994

By expanding it you get the values of a,b,c,d,e, and f as 10,-33,59,-2,-93 and 11 respectively. Put these values in the given expression and you get the answer as 994.

The expansion above can change into 10x^5 - 33x^4 + 59x^3 -2x^2 -93x + 11. That, a= 10, b=-33, c=59, d=-2, e=-93, f=11. So, 211a + 65b + 19c + 5d + e + f = 211.10 + (-33)65 + 19.59 + (-2)5 + (-93) + 11 = 994. ANSWER : \box_994