यदि `x^(3)-3x^(2)+3x+7=(x+1)(ax^(2)+bx+c)`, तो a + b + c =

If x^(3)-3x^(2)+3x-7=(x+1)(ax^(2)+bx+c) then a+b+c=4(b)12(c)-10(d)3

If x^3-3x^2+3x+7=(x+1)(a x^2+b x+c),\ then a+b+c= (a)4 (b) 12 (c) -10\ (d) 3

(7x^(3)+3x^(2)-x+1)/(x+1)=(ax^(2)+bx+c)-(2)/(x+1)

Differentiate (i) 6ax^(5)+ 4x^(3) -3x^(2) +2x -7 (ii) 5x^(-3//2) + 4/sqrtx + sqrtx -7/x (iii) ax^(3) + bx^(2) + cx+ d, where a,b,c,d are constants

If 3x^(5)-7x^(4)+4x^(3)+ax^(2)+bx+c=0 is a reciprocal equation of second type then (a,b,c)=

If f(x) =X^(3) + ax^(2) +bx +c has extreme values at x=-1 and x=3 . Find a,b,c.