How do find the quotient of (10b^2 + b - 1)/(2b + 3)10b2+b12b+3?

1 Answer
Aug 6, 2018

(10b^2+b-1)/(2b+3)=(10b^2+15b-14b-21+20)/(2b+3)10b2+b12b+3=10b2+15b14b21+202b+3
=(5b(2b+3)-7(2b+3)+20)/(2b+3)=(5b-7)+20/(2b+3)=5b(2b+3)7(2b+3)+202b+3=(5b7)+202b+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