what should be subtracted from 5y^5-3y^3-6y^2 so that(9y^3-5y) becomes its factor
Answers
Answered by
1
9y^3-5y)5y^5-3y^3-6y^2(5/9y^2-2/81
5y^5-25/9y^3
==========================
-2/9y^3-6y^2
-2/9y^3+10/81y
======================
-6y^2-10/81y
hence
-y(6y+10/81) is remaining when 9y^3-5y
is a factor of above expression
5y^5-25/9y^3
==========================
-2/9y^3-6y^2
-2/9y^3+10/81y
======================
-6y^2-10/81y
hence
-y(6y+10/81) is remaining when 9y^3-5y
is a factor of above expression
savleen:
please write in correct form
Similar questions