How do you divide #(2v^3-20v^2+56v-46)div(v-6)# using synthetic division?

2 Answers
Jun 11, 2017

Answer:

The remainder is #=2# and the quotient is #=2v^2-8v+8#

Explanation:

Let's perform the synthetic division

#color(white)(aaaa)##6##color(white)(aaaaa)##|##color(white)(aaaa)##2##color(white)(aaaaaa)##-20##color(white)(aaaaaa)##56##color(white)(aaaaa)##-46#
#color(white)(aaaaaaaaaaaa)#_________

#color(white)(aaaa)##color(white)(aaaaaaa)##|##color(white)(aaaa)##color(white)(aaaaaaaa)##12##color(white)(aaaa)##-48##color(white)(aaaaaaa)##48#
#color(white)(aaaaaaaaaaaa)#________

#color(white)(aaaa)##color(white)(aaaaaaa)##|##color(white)(aaaa)##2##color(white)(aaaaaa)##-8##color(white)(aaaaaaa)##8##color(white)(aaaaaaaa)##color(red)(2)#

The remainder is #=2# and the quotient is #=2v^2-8v+8#

#(2v^3-20v^2+56v-46)/(v-6)=2v^2-8v+8+2/(v-6)#

Jun 11, 2017

Answer:

#2v^2 -8v+8 " remainder " 2#

Or #" "2v^2 -8v+8 + 2/(v-6)#

Explanation:

In synthetic division, you use only the coefficients, making sure that all powers of #v# are considered:

#color(red)(2)v^3 color(red)(-20)v^2 color(red)(+56)v^1 color(red)(-46)v^0#

Make #v-6=0" "rarr v = color(blue)(6)" "larr# this is written 'outside'.

#color(white)(xxx.x)|color(red)(2) " "color(red)(-20)" " color(red)(+56)" " color(red)(-46)#
#color(white)(xx.xx)|darr#
#color(white)(xx.x)color(blue)(6)|ul(color(white)(xxxxx)12" "-48" "48#
#color(white)(xxxxxx)color(red)(2)" "-8" "+8" "2larr# the remainder

Steps are as follows:

Bring down the #color(red)(2)#

Multiply #color(blue)(6) xx color(red)(2) = 12#

Add #-20+12 = -8#

Multiply #color(blue)(6) xx -8 = -48#

Add #+56-48 = +8#

Multiply #color(blue)(6) xx 8 =48#

Add #-46+48 =+2#

What do the numbers tell us?

#2v^3div v= 2v^2" "larr# this is the first term in the quotient.

The numbers represent the co-efficients of the following terms:

#" "color(red)(2)" "-8" "+8" "2larr# the remainder

#" "2v^2" "-8v" "+8" remainder" 2#

This is the quotient.