# factoring 3rd order polynomial...

• Oct 13th 2008, 09:51 AM
jonbob81
factoring 3rd order polynomial...
hey can someone remind me how you get the factors of something like this:

-x^3 + 5x^2 - 5x - 24 = 0

thank you for your help, its been so long since idone these i tried googling but could not find a systematic approach,
• Oct 14th 2008, 12:51 AM
mr fantastic
Quote:

Originally Posted by jonbob81
hey can someone remind me how you get the factors of something like this:

-x^3 + 5x^2 - 5x - 24 = 0

thank you for your help, its been so long since idone these i tried googling but could not find a systematic approach,

As far as I can see -x^3 + 5x^2 - 5x - 24 has no 'simple' factors. A general approach would be to use the cubic formula, but often that's like killing a fly with an elephant gun. Sometimes grouping will work, sometimes simple linear factors can be found by inspection.
• Oct 14th 2008, 01:23 AM
jonbob81
i was thinking the same. i was trying to produce the eigenvalues of this matrix

call lambda x

(2-x ) 0 3
0 (6-x) -3
3 -3 (2-x)

and that formula below was what i got
• Oct 14th 2008, 02:36 AM
CaptainBlack
Quote:

Originally Posted by jonbob81
hey can someone remind me how you get the factors of something like this:

-x^3 + 5x^2 - 5x - 24 = 0

thank you for your help, its been so long since idone these i tried googling but could not find a systematic approach,

If this has rational roots (and so nice factors) by the rational root theorem the roots are amoung the factors of 24, so try +/11, +/-2, +/-3, +/-4, +/-6, +/-8, +/-12. If none of these are roots then it has no rational roots and so no linear factors with integer coefficients.

(As it happens none of these are roots and so this cubic has no nice factors, but it does have exactly one real root)

You can find the root algebraicaly by using the Cardano-Tartaglia cubic formula, though you will probably be better off solving for the real root numerically

CB
• Oct 14th 2008, 05:25 AM
mr fantastic
Quote:

Originally Posted by jonbob81
i was thinking the same. i was trying to produce the eigenvalues of this matrix

call lambda x

(2-x ) 0 3
0 (6-x) -3
3 -3 (2-x)

and that formula below was what i got

Quote:

Originally Posted by jonbob81
[snip]
-x^3 + 5x^2 - 5x - 24 = 0

[snip]

Your formula is wrong. The determinant is (2 - x)[(6 - x)(2 - x) - 9] + 3[-3(6 - x)] = -x^3 + 10x^2 - 10x - 48, which factorises as -(x - 8)(x^2 - 2x - 6).

Solving -(x - 8)(x^2 - 2x - 6) = 0 is simple.
• Oct 14th 2008, 05:32 AM
jonbob81
Quote:

Originally Posted by mr fantastic
Your formula is wrong. The determinant is (2 - x)[(6 - x)(2 - x) - 9] + 3[-3(6 - x)] = -x^3 + 10x^2 - 10x - 48, which factorises as -(x - 8)(x^2 - 2x - 6).

Solving -(x - 8)(x^2 - 2x - 6) = 0 is simple.

hey thanks for that, i realised that later and forgot to change it ...silly me (Doh)