1. ## Solving cubic polynomials

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

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

2. Originally Posted by geft
How do you solve $-3x^3 + 7x^2 -14x + 8 = 0$?

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.

3. It shouldn't be that complicated since I wasn't taught that method. Here's the original equation.

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

4. Originally Posted by geft
How do you solve $-3x^3 + 7x^2 -14x + 8 = 0$?

It's a characteristic polynomial. Is it possible to solve it from here or must I solve them without opening too many brackets?
You could try looking for "rational roots". The "rational root theorem" says that any rational roots of $ax^n+ \cdot\cdot\dot+ bx+ c= 0$ 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.

5. Originally Posted by geft
It shouldn't be that complicated since I wasn't taught that method. Here's the original equation.

$det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0$
Perhaps you should post the original matrix since it's possible your determinant and hence characteristic equation is wrong ....

6. You're right. Here's the matrix: $
\left(
\begin{array}{ccc}
3 & 1 & -1 \\
-4 & 2 & 2 \\
-2 & 2 & 2
\end{array}
\right)$

7. Originally Posted by geft
It shouldn't be that complicated since I wasn't taught that method. Here's the original equation.

$det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0$
Originally Posted by geft
You're right. Here's the matrix: $
\left(
\begin{array}{ccc}
3 & 1 & -1 \\
-4 & 2 & 2 \\
-2 & 2 & 2
\end{array}
\right)$
You've made a couple or errors in your calculation. It should be:

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

Go back and check your work.

8. $
\left(
\begin{array}{ccc}
3 - \lambda & 1 & -1 \\
-4 & 2 - \lambda & 2 \\
-2 & 2 & 2 - \lambda
\end{array}
\right)$

$(3 - \lambda) \left|
\begin{array}{cc} 2 - \lambda & 2 \\
2 & 2 - \lambda
\end{array}
\right| + 4 \left|
\begin{array}{cc} 1 & -1 \\
2 & 2 - \lambda
\end{array}
\right| - 2 \left|
\begin{array}{cc} 1 & -1 \\
2 - \lambda & 2
\end{array}
\right|$

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

???

9. Originally Posted by geft
$
\left(
\begin{array}{ccc}
3 - \lambda & 1 & -1 \\
-4 & 2 - \lambda & 2 \\
-2 & 2 & 2 - \lambda
\end{array}
\right)$

$(3 - \lambda) \left|
\begin{array}{cc} 2 - \lambda & 2 \\
2 & 2 - \lambda
\end{array}
\right| + 4 \left|
\begin{array}{cc} 1 & -1 \\
2 & 2 - \lambda
\end{array}
\right| - 2 \left|
\begin{array}{cc} 1 & -1 \\
2 - \lambda & 2
\end{array}
\right|$

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

???

OK, I expanded along the top row. Our expresions are equivalent. And give the correct eigenvalues. So your mistake will be somewhere in the simplifying of the expression to a cubic equation. That's where you need to check.

10. Wow, somehow I got it.

$(3-x)((2-x)(2-x)-4)+4((2-x)+2)-2(4-x)$
$(3-x)(4-4x+x^2-4)+4(4-x)-2(4-x)$
$(3-x)(-4x+x^2)+(4-2)(4-x)$
$(3-x)(-x)(4-x)+2(4-x)$
$(-3x+x^2)(4-x)+2(4-x)$
$(x^2-3x+2)(4-x)$
$(x-2)(x-1)(4-x)=0$