Question 78471
We know that 

{{{f(x)/(x-3)=x^2+3x-5+2/(x-3)}}} "The polynomial f(x) divided x-3 results in a quotient of x^2+3x-5 with a remainder of 2"

So multiply both sides by {{{x-3}}} to solve for f(x)

{{{cross(x-3)(f(x)/cross(x-3))=(x-3)(x^2+3x-5+2/(x-3))}}}


{{{f(x)=(x-3)(x^2+3x-5+2/(x-3))}}}

Now let y=x-3 to make multiplication easier

{{{f(x)=y(x^2+3x-5+2/y)}}}

{{{f(x)=x^2y+3xy-5y+2*y/y)}}} Distribute

{{{f(x)=x^2y+3xy-5y+2*cross(y)/cross(y))}}} Notice the y's cancel on the last term

{{{f(x)=x^2(x-3)+3x(x-3)-5(x-3)+2}}} Now replace y with x-3

{{{f(x)=x^2*x-x^3*3+3x*x-3x*3-5*x-5*3+2}}} Distribute 

{{{f(x)=x^3-3x^2+3x^2-9x-5x+15+2}}}

{{{f(x)=x^3-14x+17}}} Combine like terms


So our function f(x) is {{{f(x)=x^3-14x+17}}}. Notice if we divide f(x) by x-3 we get our original problem:

{{{f(x)/(x-3)=(x^3-14x+17)/(x-3)=x^2+3x-5+2/(x-3)}}}


Now since we know the function is {{{f(x)=x^3-14x+17}}} then f(3) is 


{{{f(3)=(3)^3-14(3)+17=27-42+17=2}}}

So in other words, f(3)=2. So the answer is c)