How do find the quotient of #(10b^2 + b - 1)/(2b + 3)#?

1 Answer
Aug 6, 2018

Answer:

#(10b^2+b-1)/(2b+3)=(10b^2+15b-14b-21+20)/(2b+3)#
#=(5b(2b+3)-7(2b+3)+20)/(2b+3)=(5b-7)+20/(2b+3)#
#:."quotient polynomial"=q(b)=5b-7#

Explanation:

Using synthetic division :

#diamond(10b^2+b-1)div(2b+3)#

We have , #p(b)=10b^2+b-1 and "divisor :"b=-3/2#

We take ,coefficients of #p(b) to 10,1,-1#

. #-3/2 |# #10color(white)(........)1color(white)(......)-1#
#ulcolor(white)(.......)|# #ul(0color(white)( .......)-15color(white)(......)21#
#color(white)(........)10color(white)(......)-14color(white)(......)color(violet)(ul|20|#
Hence ,

#(10b^2+b-1)=(b+3/2)(10b-14)+(20)#

#(10b^2+b-1)=1/2(2b+3)(10b-14)+(20)#

#(10b^2+b-1)=(2b+3)(5b-7)+(20)#

We can see that , quotient polynomial :

#q(b)=1/2(10b-14)=5b-7 and"the Remainder"=20#