It supposed to be , the division is incorrect...
What are the roots of the cubic equcation x^{3}-8x-3=0?
Ans: D. x=-2.62, -0.38, 3
1. step
x^{3}-8x-3 = x^{2}+3x-1 By inspection +3 is a root, and x-3 is a factor. Factor out (x-3)
x-3
Please do whole the problem. How did they get x^{2}+3x-1=0
We know that (x-3)( )=x^3-8x+3 We want to find the empty bracket. It must start with x^2 and end in -1
So (x-3)(x^2 -1)=X^3-8x+3. We notice that there is no x^2 term in the answer. With the brackets as we have them so far we would have -3x^2 in the answer. So we still need +3x^2. This we will have if we add +3x to the second bracket.
If you know how to do long division you could get this from (x^3-8x-3) divided by (x-3)
So x-3=0 or x^2+3x+1= 0 which you solve using the quadratic equation formula.