How do you evaluate `f(x)=2x^3-x^4+5x^2-x` at x=3 using direct substitution and synthetic division?

Direct substitution is probably the easiest to understand. All we're doing is plugging in the `x`-value of `color(blue)3` into `f(x)`. In other words:
If

`f(x)=2x^3-x^4+5x^2-x`

then

`f(color(blue)3)=2(color(blue)3)^3-(color(blue)3)^4+5(color(blue)3)^2-color(blue)3`
`color(white)(f(3))=2(27)-" "81" "+" "5(9)" "-3`
`color(white)(f(3))=54-81+45-3`
`color(white)(f(3))=15`

To use synthetic division, we are going to "factor out" `(x-3)` from `f(x)`; the remainder of this division will be our answer.

First: arrange the coefficients of the polynomial in decreasing order in an "L" frame, including `0`'s where necessary:

`"       "| " -"1" "2" "5" ""-"1" "0`
`"       |"ul("                                      ")`

Next, attempt to factor out `(x-3)`: add a column, multiply the sum by 3, place the product in the next cell diagonally up; repeat:

`"    3 "| " -"1" "2" "5" ""-"1" "0`
`"       |"ul("         -3    -3     6    15")`
`"           ""-"1"   ""-"1"     "2"     "5"  |"ul(" "15)`

The remainder of `15` is our value of `f(3)`.

Bonus:

I'll admit, I don't remember learning to evaluate polynomials with synthetic division when I was in high school, but I can see why this works. The above synthetic division yields the following "factored" version of `f(x)`:

`f(x)=(x-3)("-"x^3-x^2+2x+5+15/(x-3))`

This can be (partially) re-distributed as:

`f(x)=(x-3)("-"x^3-x^2+2x+5)+(15(x-3))/(x-3)`

`color(white)(f(x))=(x-3)("-"x^3-x^2+2x+5)+15`

(Cancelling the `(x-3)`'s in the last term is okay, because `x=3` is in the domain of the original `f(x)`).

From here, it is easy to see that when `color(red)(x=3)`, we get

`f(color(red)3)=(color(red)3-3)(..." "..." "..." "...)+15`
`color(white)(f(3))=("    "0"    ")(..." "..." "..." "...)+15`
`color(white)(f(3))=15`.