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

1 Answer
Jan 10, 2017

Answer:

#f(x)=15#.

Explanation:

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#.