How do you use synthetic substitution to find #P(2)# for the polynomial #P(x) = 2x^3-5x^2+x+2#?

1 Answer
Jason K.
Sep 29, 2017

#P(2) = 0#

Explanation:

Using synthetic division to find the value of a function at a value is a quick and easy manner to evaluate larger polynomials, and is often used in computer programs because of how it operates using simple multiplication and addition.

Begin by making a workspace on paper where you write down, in separated columns, the coefficients of each term. You do not need to write the #x# variables, as they are implied to be there. For this problem, you begin with something which looks like this:

#" "2" "-5" "1" "2#

A note before continuing:

  • You must write down coefficients for every #x# term in descending order of power, even if a term is missing. In those cases, use 0 as the coefficient. (This doesn't apply to this problem, but it's worth noting.)

Next, you write the value of #x# you are evaluating at (in this case, 2) in a little "boxed off" area to the left of this row of numbers you just wrote. This is analogous to how you write long division in grade school. For example:

#2__|" "2" "-5" "1" "2#

Leave a little vertical space (where you will write some numbers soon), and draw a horizontal line, much like you would if this were an addition problem.

#2__|" "2" "-5" "1" "2#

#" " ul(" ")#

Lastly, copy down the first number in the row you first wrote (2) and write it underneath the line:

#2__|" "2" "-5" "1" "2#

#" " ul(" ")#
#" "2#

From now on we will repeat two steps over and over until we run out of numbers to work with:

1) Multiply the last number you wrote below the line by the number in "the box" (2 in our case), and write that answer just above the line underneath the next number to the right on the top row.

2) Add the next number to the right on the top row to the number you just copied, and write the sum under the line.

For our problem, we take the last number we wrote under the line (2), and multiply it by the number in "the box" (2) to get a product of 4. We write that in the next space, which I've colored blue in the next picture.

#2__|" "2" "-5" "1" "2#

#" "color(blue)(4)#
#" " ul(" ")#
#" "2#

Now we add up and down in that column (#-5 + 4#) and write the sum under the line (colored in blue in the next picture).

#2__|" "2" "-5" "1" "2#

#" "4#
#" " ul(" ")#
#" "2" "color(blue)(-1)#

We still have numbers in the top row to work with (1 and 2), so repeat these two steps: Multiply over, add down.

#2__|" "2" "-5" "1" "2#

#" "4" "color(blue)(-2)" "color(green)(-2)#
#" " ul(" ")#
#" "2" "-1" "color(blue)(-1)" "color(red)(0)#

The final value we wrote below the line (0 in this case and marked red) is the final value of #f(2)#.

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