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

##### 1 Answer

#### Explanation:

Direct substitution is probably the easiest to understand. All we're doing is plugging in the

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" *remainder* of this division will be our answer.

First: arrange the coefficients of the polynomial in decreasing order in an "L" frame, including

Next, attempt to factor out

The remainder of

## 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)=(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

#f(color(red)3)=(color(red)3-3)(..." "..." "..." "...)+15#

#color(white)(f(3))=(" "0" ")(..." "..." "..." "...)+15#

#color(white)(f(3))=15# .