How do you divide #(2x^3-11x^2+13x-44)div(x-5)# using synthetic division?

1 Answer
Mar 8, 2018

Answer:

#2x^2-1x+8 color(white)("xxx")(-4)/(x-5)#

Explanation:

Note:
#color(white)("XXX")color(magenta)5# (used below) is the value of #x# required to make the divisor, #(x-5)# equal to zero.

rows [0] and [4] are not really part of synthetic division; I include them for reference purposes only.

row [1] are the coefficients of the powers of #x# (when the expression is formed in standard notation and using #0# for any missing powers of #x#).

row [2] contains, for each column, the product of #color(magenta)5# (see above) and the value from the previous column of row [3].

row [3] contains, for each column, the sum of the values in the same column for rows [1] and [2]

The notations used in the (optional) row 4 are the powers of #x# from the (optional) row [1] reduced by 1, with the final column, marked as #color(gray)("R")#, the remainder.

#{: ([0],," | ",color(gray)(x^3),color(white)("x")color(gray)(x^2),color(white)("x")color(gray)(x^1),color(white)("x")color(gray)(x^0)), ([1],," | ",2,-11,+13,-44), ([2],ul(+color(white)("xx"))," | ",ul(color(white)("xxxx")),ul(color(white)("x")10),ul(color(white)("x")-5),ul(color(white)("x")40)), ([3],xx color(magenta)5," | ",color(white)("x")2,color(white)("x")-1,color(white)("xxx")8,color(white)("x")-4), ([4],," | ",color(gray)(x^2),color(white)("x")color(gray)(x^1),color(white)("xx")color(gray)(x^0),color(white)("xx")color(gray)R) :}#

Normally when performing synthetic division, only rows [1], [2], and [3] are required.