" Facturise "rArr2y^(3)+y^(2)-2y-1

Factorise : - 2y^(3)+y^(2)-2y-1

Factorise : 2y^3+y^2-2y-1 .

Find the equation of a circle, the co-ordinates of the ends of whose diameter are (-1,-3) and (2,5) Now the equation of circle (x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2))=0 rArr(x+1)(x-2)+(y+3)(y-5)=0 rArr x^(2)-x-2+ y^(2)-2y-15=0 rArr x^(2)+y^(2)-x-2y-17=0 We is the required equation.

((y^(3)-3y^(2)+5y-1)/(y-1))

If y=(sqrt(3)+1)/(2) then find the value of 4y^(3)+2y^(2)-8y+10=

The value of lim_(x rarr1)(y^(3))/(x^(3)-y^(2)-1) as (x,y)rarr(1,0) along the line y=x-1 is