p(x)=x^3 -6x^2 +13x -10. find all the solutions to the equation p(x)=0.

Question

i am not allowed to use -b/2A

Answer 1

of course -b/2a does not work. It's not a quadratic.

Furthermore, x = -b/2a gives the location of the vertex of a quadratic, not the roots.

You know that any rational roots must be among ±1,±2,±5,±10

A little synthetic division shows that

p(x) = (x-2)(x^2-4x+5)

now use the quadratic formula to get the complex roots.