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## Homework Statement

Find the real solutions of [itex]x^3+6x^2-8x+1[/itex]

## Homework Equations

quadratic

## 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

[tex]

x^3+6x^2-8x+1=(x-1)(Ax^2+Bx+C) \\

x^3+6x^2-8x+1=Ax^3+Bx^2+Cx-Ax^2-Bx-C \\

[/tex]

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

[itex]\frac{-7\pm\sqrt{53}}{2}[/itex]

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.