Question #b6e0c

The result that corresponds to

#(f(x))/(x-c) = q(x) + (r(x))/(x-c)#

is:

#(x^5 + 2x^2 - x + 2)/(x+1) = stackrel(q(x))overbrace(x^4 - x^3 + x^2 + x - 2) + stackrel(r(x)"/"(x-c))overbrace(4/(x+1))#,

where #q(x)# is the solution disregarding the remainder, and #r(x)# is the remainder.

The actual remainder is therefore #r(x) = 4#.

The remainder theorem states that the remainder of a division of #f(x)# by #x-c# is #f(c)#. In other words, for this problem, since #x-c = x+1#, we have that #c = -1#, and so, the remainder is expected to be:

#f(c) = f(-1) = (-1)^5 + 2(-1)^2 - (-1) + 2#

#= -1 + 2 + 1 + 2#

#= 4#

Let's see if we get that, because that tells us whether we did our division correctly.

Synthetic division can be written as:

#ul(-1) | 1" "0" "0" "2" "-1" "2#
#" "" "" "#
#" "" ""--------------------------------"#

where each coefficient is the coefficient for each degree in the polynomial, and the number in the upper-left box is #c#. The #x#'s are implied. For example, since #x^3 + 7x^2 - 2# is also

#color(green)(1)x^3 + color(green)(7)x^2 + color(green)(0)x color(green)(-2)#,

it is written as simply "#1" "7" "0" "-2#".

The general steps are:

1. Move the first coefficient down.
2. Multiply the divisor factor (the #c# in #x-c#) by the coefficient currently below the horizontal line. Place the result below the next coefficient, above the line.
3. Add the latest column and place the result below the line.
4. Repeat steps 2 and 3 until you've reached the farthest bottom-right. That is your remainder.

Following each step (which is repetitive), we get:

Bring the coefficient down.

#ul(-1) | 1" "0" "0" "2" "-1" "2#
#" "" "" "#
#" "" ""--------------------------------"#
#" "" "1#

Multiply by #-1# and place on next column; add up.

#ul(-1) | 1" "0" "0" "2" "-1" "2#
#" "" "" "-1#
#" "" ""--------------------------------"#
#" "" "1color(white)(.)-1#

Continue to next column:

#ul(-1) | 1" "0" "0" "2" "-1" "2#
#" "" "" "-1color(white)(..)1#
#" "" ""--------------------------------"#
#" "" "1color(white)(.)-1color(white)(A)1#

Continue to next column:

#ul(-1) | 1" "0" "0" "2" "-1" "2#
#" "" "" "-1color(white)(..)1color(white)(.)-1#
#" "" ""--------------------------------"#
#" "" "1color(white)(.)-1color(white)(A)1color(white)(..)1#

Continue to next column:

#ul(-1) | 1" "0" "0" "2" "-1" "2#
#" "" "" "-1color(white)(..)1color(white)(.)-1color(white)(.)-1#
#" "" ""--------------------------------"#
#" "" "1color(white)(.)-1color(white)(A)1color(white)(..)1color(white)(..)-2#

Last column!

#ul(-1) | 1" "0" "0" "2" "-1" "2#
#" "" "" "-1color(white)(..)1color(white)(.)-1color(white)(.)-1color(white)(..)2#
#" "" ""--------------------------------"#
#" "" "1color(white)(.)-1color(white)(A)1color(white)(..)1color(white)(..)-2color(white)(..)bb(4)#

You can see that we do indeed get a remainder of #4#, as anticipated from evaluating #f(-1)# initially.

Rewriting the answer in normal form, we get:

#(x^5 + 2x^2 - x + 2)/(x+1) = color(blue)(stackrel(q(x))overbrace(x^4 - x^3 + x^2 + x - 2) + stackrel(r(x)"/"(x-c))overbrace(4/(x+1)))#

CHALLENGE: Try repeating this problem with #x^5 + 2x^2 - x - 2#. See if you get no remainder.