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?
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?
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.
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.
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.
$\displaystyle
\left(
\begin{array}{ccc}
3 - \lambda & 1 & -1 \\
-4 & 2 - \lambda & 2 \\
-2 & 2 & 2 - \lambda
\end{array}
\right)$
$\displaystyle (3 - \lambda) \left|
\begin{array}{cc} 2 - \lambda & 2 \\
2 & 2 - \lambda
\end{array}
\right| + 4 \left|
\begin{array}{cc} 1 & -1 \\
2 & 2 - \lambda
\end{array}
\right| - 2 \left|
\begin{array}{cc} 1 & -1 \\
2 - \lambda & 2
\end{array}
\right|$
$\displaystyle det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0$
???
Wow, somehow I got it.
$\displaystyle (3-x)((2-x)(2-x)-4)+4((2-x)+2)-2(4-x)$
$\displaystyle (3-x)(4-4x+x^2-4)+4(4-x)-2(4-x)$
$\displaystyle (3-x)(-4x+x^2)+(4-2)(4-x)$
$\displaystyle (3-x)(-x)(4-x)+2(4-x)$
$\displaystyle (-3x+x^2)(4-x)+2(4-x)$
$\displaystyle (x^2-3x+2)(4-x)$
$\displaystyle (x-2)(x-1)(4-x)=0$