How do you divide #(-18a+3a^3+9+6a^2)div(-3+3a)# using synthetic division?

2 Answers
Aug 14, 2018

Answer:

The remainder is #0# and the quotient is #=a^2+3a-3#

Explanation:

Divide the dividend and the divisor by #3#

Let's perform the synthetic division

#color(white)(aaaa)##1##|##color(white)(aaaa)##1##color(white)(aaaa)##2##color(white)(aaaaaa)##-6##color(white)(aaaaa)##3#

#color(white)(aaaaa)##|##color(white)(aaaa)##color(white)(aaaaa)##1##color(white)(aaaaaaaa)##3##color(white)(aaa)##-3#

#color(white)(aaaaaaaaa)###_________

#color(white)(aaaaaa)##|##color(white)(aaaa)##1##color(white)(aaaa)##3##color(white)(aaaaa)##-3##color(white)(aaaaa)##color(red)(0)#

The remainder is #0# and the quotient is #=a^2+3a-3#

#(a^3+2a^2-6a+3)/(a-1)=a^2+3a-3#

Aug 14, 2018

Answer:

#(3a^3+6a^2-18a+9)div(3a-3)=(a^2+3a-3 )+(0)#

Explanation:

Here ,

#(-18a+3a^3+9+6a^2)div(-3+3a)#

#=>(3a^3+6a^2-18a+9)/(3a-3)=(cancel3(a^3+2a^2-6a+3))/(cancel3(a-1)#

#=>(a^3+2a^2-6a+3)div(a-1)#

Using synthetic division :

We have , #p(a)=(a^3+2a^2-6a+3) and "divisor : " a=1#

We take , coefficients of #p(a) to 1,2,-6,3#

. #1|# #1color(white)(........)2color(white)(......)-6color(white)(........)3#
#ulcolor(white)(...)|# #ul(0color(white)( ........)1color(white)(..........)3color(white)(.....)-3#
#color(white)(......)1color(white)(........)3color(white)(....)-3color(white)(........)color(violet)(ul|0|#
We can see that , quotient polynomial :

#q(a)=a^2+3a-3 and"the Remainder"=0#

Hence ,

#(a^3+2a^2-6a+3)div(a-1)=(a^2+3a-3 )+(0)#

Multiplying numerator and denominator of LHS by #3#

#(3(a^3+2a^2-6a+3))/(3(a-1))=(a^2+3a-3 )+(0)#

#(3a^3+6a^2-18a+9)/(3a-3)=(a^2+3a-3 )+(0)#