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.
p(x)=x^3 -6x^2 +13x -10. find all the solutions to the equation p(x)=0.
i am not allowed to use -b/2A
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