Finding solutions to a cubic

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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 dont 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.
 

Answers and Replies

  • #2
Office_Shredder
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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 [itex] x=1,\ x=\frac{-7+\sqrt{53}}{2}[/itex] and [itex] x=\frac{-7-\sqrt{53}}{2}[/itex]
 
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  • #3
LCKurtz
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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]
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...
 

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