How do you solve $\displaystyle -3x^3 + 7x^2 -14x + 8 = 0$?

It's a characteristic polynomial. Is it possible to solve it from here or must I solve them without opening too many brackets?

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- Oct 25th 2009, 09:43 PMgeftSolving cubic polynomials
How do you solve $\displaystyle -3x^3 + 7x^2 -14x + 8 = 0$?

It's a characteristic polynomial. Is it possible to solve it from here or must I solve them without opening too many brackets? - Oct 25th 2009, 10:47 PMearboth
The LHS of the equation can't be factored. Therefore you can use either the Cardanic formula (see here: Cubic function - Wikipedia, the free encyclopedia) or an iterative method like Newton's method to get an approximate solution.

- Oct 25th 2009, 11:00 PMgeft
It shouldn't be that complicated since I wasn't taught that method. Here's the original equation.

$\displaystyle det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0$ - Oct 26th 2009, 03:40 AMHallsofIvy
You could try looking for "rational roots". The "rational root theorem" says that any rational roots of $\displaystyle ax^n+ \cdot\cdot\dot+ bx+ c= 0$ are of the form x= r/s where s is an integer factor of a, the leading coefficient, and s is an integer factor of c, the constant term.

Here, the leading coefficient is "3" which has factors 1, -1, 3, and -3. The constant term is "8" which has factors 1, -1, 2, -2, 4, -4, 8, and -8. The only possible rational roots are 1, -1, 2, -2, 4, -4, 8, -8, 1/3, -1/3,, 2/3, -2/3, 4/3, -4/3, 8/3, and -8/3. Trying each of those will either give you a root or show that this equation has no rational roots. If there are no rational roots there will not be any solution simpler than Cardano's. - Oct 26th 2009, 03:56 AMmr fantastic
- Oct 26th 2009, 06:54 AMgeft
You're right. Here's the matrix: $\displaystyle

\left(

\begin{array}{ccc}

3 & 1 & -1 \\

-4 & 2 & 2 \\

-2 & 2 & 2

\end{array}

\right) $ - Oct 26th 2009, 06:02 PMmr fantastic
You've made a couple or errors in your calculation. It should be:

$\displaystyle det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) {\color{red}- 1}] - 2[{\color{red}-2} - \lambda] = 0$.

Go back and check your work. - Oct 26th 2009, 09:39 PMgeft
$\displaystyle

\left(

\begin{array}{ccc}

3 - \lambda & 1 & -1 \\

-4 & 2 - \lambda & 2 \\

-2 & 2 & 2 - \lambda

\end{array}

\right)$

$\displaystyle (3 - \lambda) \left|

\begin{array}{cc} 2 - \lambda & 2 \\

2 & 2 - \lambda

\end{array}

\right| + 4 \left|

\begin{array}{cc} 1 & -1 \\

2 & 2 - \lambda

\end{array}

\right| - 2 \left|

\begin{array}{cc} 1 & -1 \\

2 - \lambda & 2

\end{array}

\right|$

$\displaystyle det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0$

???

- Oct 26th 2009, 11:43 PMmr fantastic
- Oct 26th 2009, 11:55 PMgeft
Wow, somehow I got it.

$\displaystyle (3-x)((2-x)(2-x)-4)+4((2-x)+2)-2(4-x)$

$\displaystyle (3-x)(4-4x+x^2-4)+4(4-x)-2(4-x)$

$\displaystyle (3-x)(-4x+x^2)+(4-2)(4-x)$

$\displaystyle (3-x)(-x)(4-x)+2(4-x)$

$\displaystyle (-3x+x^2)(4-x)+2(4-x)$

$\displaystyle (x^2-3x+2)(4-x)$

$\displaystyle (x-2)(x-1)(4-x)=0$