Could someone help me with this?
It doesn't seem difficult, but my answer doesn't suit to the answer in the book
so basically x^3 - 6x - 9 = 0
We can factor out (x-3) from here, leaving
(x-3)(x^2+3*x+3) = 0
x-3 = 0, so x=3 is a solution
x^2 + 3x + 3 = 0
This can be solved easily using the quadratic formula
thanx man, but I did it in another more complicated way.
D<0 no solutions
Now the answer is correct.
Could u explain me how you get this (x^2+3*x+3)?
Last edited by Rose-Red (2006-01-08 09:21:57)
God factored out x-3 from the original equation. (x-3)(x²+3x+3) = x³-6x-9
but i dont understand how it is possible. I cant find any rule or formula that allows to do that
By the rational roots theorem, the only possible rational solutions are -9 -3 -1 1 3 and 9
By the rule of signs, there is one sign change, so there is a positive root, so start with positive numbers.
After finding that 3 is a root, we know that we can factor out (x-3)
Use synthetic division to find out the remaining quadratic: