The sum of the reciprocals of the roots of x^3+ax^2+bx+c=0 is:?

A) a/b B)a/c C) -a/c D) -b/c E) 1/ (a+c)

The sum of the reciprocals of the roots of x^3+ax^2+bx+c=0 is:?
Label the roots of the polynomial f(x) as u,v, and w. Then the coefficients a,b, and c are, up to a minus sign, the elementary symmetric functions in terms of the roots u,v, and w. That is, a = -(u+v+w), b = uv+uw+vw, and c = -uvw. Therefore, the sum of the reciprocals of the roots is just -b/c.
Reply:Do your own homework, we can't sit the test for you.
Reply:Suppose that R is a root, then R^3 + aR^2 + bR + c = 0. Divide by R^3. Therefore 1 + a/R + b/R^2 + c/R^3 = 0, showing that 1/R is a root of cx^3 + bx^2 + ax + 1 = 0, hence that all the roots of cx^3 + bx^2 + ax + 1 = 0 are the reciprocals of all the roots of x^3 + ax^2 + bx + c = 0.





The sum of the roots of cx^3 + bx^2 + ax + 1 = 0 is -b/c, which you already knew, I'm sure.