Simple factor of equation I have this polynomial: $5z^4-12z^3+30z^2-12z+5$ 
How do I factor it to get the following?: $(5z^2-2z+1)(z^2-2z+5)$
Can someone show me the procedure to perform whenever I encounter with a case like this? Thank you.

You should exploit the symmetry: write the polynomial as
$$
z^2\left(5z^2+\frac{5}{z^2}-12z-\frac{12}{z}-30\right)
$$
and observe that
$$
z^2+\frac{1}{z^2}=\left(z+\frac{1}{z}\right)^{\!2}-2
$$
so you can rewrite the expression as
$$
z^2\left(5\left(z+\frac{1}{z}\right)^{\!2}-12\left(z+\frac{1}{z}\right)-40\right)
$$
The polynomial
$$
5t^2-12t-40
$$
has roots
$$
a=\frac{6+\sqrt{236}}{5},\quad b=\frac{6-\sqrt{236}}{5}
$$
so we get
$$
5z^2\left(z+\frac{1}{z}-a\right)\left(z+\frac{1}{z}-b\right)
$$
that can be rewritten as
$$
5(z^2-az+1)(z^2-bz+1)
$$

If your polynomial is
$$
5z^4-12z^3\color{red}{+}30z^2-12z+5
$$
the same procedure would give a polynomial in $t$ without real roots. In particular, there is no real root for the polynomial.
In this case you know that if $\alpha$ is a root, also $\alpha^{-1}$ is  root. Since a real factorization exists, the roots must have modulus $1$ and if we pair the conjugate pairs, we get a factorization in the form
$$
5z^4-12z^3+30z^2-12z+5=
(5z^2+az+b)(bz^2+az+5)
$$
(try seeing why). Now it's quite easy to find $a$ and $b$.
Or you can try finding the complex roots. The procedure reduces to the polynomial
$$
5t^2-12t+20
$$
whose roots are
$$
\frac{6+8i}{5},\qquad \frac{6-8i}{5}
$$
so you just have to solve the equations
$$
z+\frac{1}{z}=\frac{6+8i}{5}
\qquad
z+\frac{1}{z}=\frac{6-8i}{5}
$$
