What are the steps to find the roots of a cubic equation?

[r-soy]

Hi,

If we have this equation:

\(\displaystyle m^3 - m^2 - 8m + 12 = 0 \)

how we can get the roots \(\displaystyle m_1,\,m_2,\,m_3\)?

can someone please help... ?

 

[Prove It]

Re: How we can know that m1,2,3

Hi

If we have this equation :
m^3 - m^2 - 8m + 12 = 0
how we can get m 1 and m2 and m3
can please help ... ?

The factors of 12 are -12, -6, -4, -3, -2, -1, 1, 2, 3, 4, 6 and 12.

Let x equal each of these values. See if any of them make the expression equal to 0. If so, then (x - that number) is a factor. Long divide the expression to get the remaining quadratic factor which you can then solve using the quadratic formula if needbe.

 

[mathmaniac1]

Re: How we can know that m1,2,3

Hey Prove It, you were lucky as the constant and the coefficient of m^2 were both integers...
Otherwise giving value of x won't help,I think...

 

[Prove It]

Re: How we can know that m1,2,3

Hey Prove It, you were lucky as the constant and the coefficient of m^2 were both integers...
Otherwise giving value of x won't help,I think...

If my method wasn't able to be used in this case I wouldn't have suggested it :) but you are correct, it does not work for all cubics.

If you're really unlucky, you might have to use the Cubic Formula.

View attachment 700

 

[mathmaniac1]

Re: How we can know that m1,2,3

If you're really unlucky, you might have to use the Cubic Formula...

...which is very nasty...

I think it would be more helpful if you explained why you looked for factors of 12.I mean the nature of zeros.I am bad at latexing (Sadface) otherwise I would have done it...

 

[MarkFL]

Re: How we can know that m1,2,3

...

I think it would be more helpful if you explained why you looked for factors of 12...

I think Prove It, being aware that the OP is posting problems in the differential equations and calculus forums, is fair in assuming that the rational roots theorem has already been covered in previous courses.

 

[mathmaniac1]

Re: How we can know that m1,2,3

I think Prove It, being aware that the OP is posting problems in the differential equations and calculus forums, is fair in assuming that the rational roots theorem has already been covered in previous courses.

Alright then...

 

[MarkFL]

If you wish to explain the use of the rational roots theorem you are certainly free to do so. I didn't mean it should not be done, only that it is fair to assume it isn't necessary, at least not for the OP. (Wink)

 

[mathmaniac1]

If you wish to explain the use of the rational roots theorem you are certainly free to do so. I didn't mean it should not be done, only that it is fair to assume it isn't necessary, at least not for the OP. (Wink)

If its not for him,for whom should I do it? and I don't want to spend hours trying to perfect my latex now...

Will it be okay if I start a new thread in Latex Help to learn latex for me?

 

[MarkFL]

If you wish to explain it for the benefit of others who read this topic, hoping to find help on how to find the roots of a cubic, that's fine.

In an effort to not take this topic too far off course, I will address your $\LaTeX$ question privately, by VM.

 

[mathmaniac1]

Let \(\displaystyle p,q \ and \ r\) be the roots of the cubic,then \(\displaystyle (x-p)(x-q)(x-r)=0\).
Simplyfying we get \(\displaystyle x^3-(p+q+r)x^2+(pq+qr+rq)x+pqr=0\)
And comparing it with the general form of a cubic ,i.e, \(\displaystyle ax^3+bx^2+cx+d=0\),we get \(\displaystyle p+q+r=\frac{-b}{a}\),(In fact this is true for equations of any degree)\(\displaystyle pq+qr+pr=\frac{c}{a}\) and \(\displaystyle pqr=\frac{d}{a}\)
Using this we can make guesses about the roots of a cubic...