Find the Characteristic root of the matrix
$\displaystyle
\begin{bmatrix}
2 & 2 & 1 \\
1 & 3 & 1 \\
1 & 2 & 2 \\
\end{bmatrix}$
Write the characteristic polynomial of the matrix. Call the matrix A, then the characteristic polynomial is $\displaystyle \text{det \(A - tI\)}$.
I take it that you are worried about the cubic nature of the polynomial and you dont know how to solve cubics.
In this particular case, observe that if you expand the determinant as:
$\displaystyle \text{det (A - tI)} = (2 - t)[(3 - t)(2 - t) - 2] - 2[(2 - t)- 1] + 1[2 - (3 - t)]$$\displaystyle = (2 - t)[t^2 - 5t + 4] - 2[1 - t] + 1[t - 1]$
Now observe that $\displaystyle t - 1$ is a factor of all the three [] terms.
$\displaystyle (2 - t)[t^2 - 5t + 4] - 2[1 - t] + 1[t - 1] = (2 - t)[(t-1)(t-4)] + 2[t - 1] + 1[t - 1]$
I hope you can finish it now
I dont know how you got the above equation. Anyway I have worked it out completely below. But I have to make a comment about the way you did the algebra above.
Why did you write $\displaystyle 3 \lambda^2 -7 \lambda + 5$ as $\displaystyle (\lambda - 1)(3 \lambda - 4) + 1$? When you want to find a root for a polynomial, you want to factor it completely. You cant leave terms like + 1 hanging in the equality.
Note: You want factor it completely because you know that the product of 2 real numbers is 0 means at least one of them is 0.
I will continue from here,
$\displaystyle (2 - t)[(t-1)(t-4)] + 2[t - 1] + 1[t - 1] =$$\displaystyle (t-1)[(2 - t)(t-4) + 3] = (t-1)[-t^2 + 6t - 5] = (t - 1)[-(t-1)(t-5)]$
Thus $\displaystyle \det(A - tI) = 0 \Rightarrow (t - 1)[-(t-1)(t-5)] = 0 \Rightarrow t = 1,1,5$