How do you divide #(3r^3+34r^2+89r+75)div(r+8)# using synthetic division?

1 Answer
Jul 18, 2018

Answer:

#(3r^3+34r^2+89r+75)=(r+8)(3r^2+10r+9)+3#

Explanation:

Here,

#p(r)=3r^3+34r^2+89r+75and # divisor #r=-8#

We take coefficients of #p(r)# and set the problem as shown below.

#color(red)("Put zero {0} below the first number {3} and add : " 3+0=3#

#(-8) |# #color(red)3color(white)(.....)34color(white)(.....)89color(white)(....)75#
#ulcolor(white)((....2))|# #ul(color(red)0color(white)(..................................)#
#color(white)(..........)color(red)3#

#color(blue)("Now multiply this {3}with divisor"# #color(blue)((-8)to3xx(-8)=-24# #color(blue)("and put below second number{34} and add"#

#color(blue)(to34+(-24)==10#

#(-8) |# #3color(white)(.....)color(blue)(34)color(white)(.....)89color(white)(....)75#
#ulcolor(white)((....2))|# #ul(0color(white)(....)color(blue)(-24)color(white)(......................)#
#color(white)(..........)3color(white)(.....)color(blue)(10color(white)(.....20color(white)(.........)ul|0|#
Again repeat the process :

#i.e. color(brown)(10xx(-8)=-80 and 89+(-80)=9#

#(-8) |# #3color(white)(.......)34color(white)(.....)color(brown)(89)color(white)(....)75#
#ulcolor(white)((....2))|# #ul(0color(white)(...)-24color(white)(..)color(brown)(-80)color(white)(.........10#
#color(white)(..........)3color(white)(.....)10color(white)(........)color(brown)(9)color(white)(.........ul|0|#

Again , #color(violet)(9xx(-8)=-72 and (75)+(-72)=3#

#(-8) |# #3color(white)(.....)34color(white)(.......)89color(white)(........)color(violet)(75#
#ulcolor(white)((....2))|# #ul(0color(white)(.)-24color(white)(..)-80color(white)(.....)color(violet)(-72#
#color(white)(..........)3color(white)(......)10color(white)(........)9color(white)(.........)color(violet)(ul|3|#
We can see that , quotient polynomial :

#q(r)=3r^2+10r+9 and"the Remainder"=3#

Hence ,

#(3r^3+34r^2+89r+75)=(r+8)(3r^2+10r+9)+3#