Find the real solutions of [itex]x^3+6x^2-8x+1[/itex]
The Attempt at a Solution
First I tried simple numbers starting with 1 and found that +1 was a solution and therefore I could write
And then equated the coeffeciants
For x^3, [itex]1=A[/itex]
For x^2, [itex]6=B-A[/itex]
For x^1, [itex]-8=C-B[/itex]
For x^0, [itex]1=-C[/itex]
With knowing that A=1 and C= -1 , B can be found to be 7 .
And therefore I can write [itex]x^3+6x^2-8x+1=(x-1)(x^2+7x-1)[/itex]
If I then use the quadratic formula on the quadratic factor I end up with
And as 53 is prime that is as far as it can go without losing accuracy.
I don't know if this is all I have to do, if I have done what I have done correctly or what. I am a bit confused. Any help/advice appreciated.