Finding solutions to a cubic

[FaraDazed]

Homework Statement

Find the real solutions of $x^3+6x^2-8x+1$

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

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

And then equated the coeffeciants

For x^3, $1=A$
For x^2, $6=B-A$
For x^1, $-8=C-B$
For x^0, $1=-C$

With knowing that A=1 and C= -1 , B can be found to be 7 .

And therefore I can write $x^3+6x^2-8x+1=(x-1)(x^2+7x-1)$

If I then use the quadratic formula on the quadratic factor I end up with
$\frac{-7\pm\sqrt{53}}{2}$

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.

 

[Office_Shredder]

That looks entirely correct, and is all you have to do. You can include a line at the end that says

Therefore, the solutions are $x=1,\ x=\frac{-7+\sqrt{53}}{2}$ and $x=\frac{-7-\sqrt{53}}{2}$

 

[LCKurtz]

Homework Statement

Find the real solutions of $x^3+6x^2-8x+1$

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

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

And then equated the coeffeciants

For x^3, $1=A$
For x^2, $6=B-A$
For x^1, $-8=C-B$
For x^0, $1=-C$

With knowing that A=1 and C= -1 , B can be found to be 7 .

And therefore I can write $x^3+6x^2-8x+1=(x-1)(x^2+7x-1)$

I'm curious why you didn't just divide the cubic by $x-1$ using either long division or, preferably, synthetic division to get the quadratic factor. Much less work...