The given polynomial is:

$f(z) = 2z^4 - 10z^3 + 26z^2 - 38z + 20$

You are provided that one of the roots is $z = 1 - 2i$. Since the coefficients of the polynomial are real, the complex conjugate of this root, $z = 1 + 2i$, must also be a root of the polynomial.

This means that two of the roots are $z = 1 - 2i$ and $z = 1 + 2i$. These two roots correspond to the quadratic factor $(z - (1 - 2i))(z - (1 + 2i))$.

We can now compute this quadratic factor:

$(z - (1 - 2i))(z - (1 + 2i)) = (z - 1)^2 - (2i)^2 = (z - 1)^2 + 4$

Expanding $(z - 1)^2$:

$(z - 1)^2 = z^2 - 2z + 1$

Thus, the quadratic factor becomes:

$z^2 - 2z + 1 + 4 = z^2 - 2z + 5$

Now, we can divide the original polynomial $f(z)$ by $z^2 - 2z + 5$ to find the other quadratic factor. I will now perform this division.The division of the original polynomial by $z^2 - 2z + 5$ gives the quotient:

$2z^2 - 6z + 4$

and a remainder of 0, which means the factorization is exact.

Now, we solve for the roots of the quadratic factor $2z^2 - 6z + 4$. Dividing the quadratic equation by 2:

$z^2 - 3z + 2 = 0$

Factoring this:

$(z - 1)(z - 2) = 0$

Thus, the other two roots are $z = 1$ and $z = 2$.

Final Answer:

The four roots of the polynomial are:

• $z = 1 - 2i$
• $z = 1 + 2i$
• $z = 1$
• $z = 2$

Let me know if you'd like further details or clarifications!

Follow-up questions:

1. How do you verify if a complex root has its conjugate as a root in real-coefficient polynomials?
2. What method can be used to perform polynomial division by hand?
3. How does the degree of a polynomial relate to the number of roots it has?
4. What is the significance of the remainder being zero in polynomial division?
5. How can the quadratic formula be used for non-factorable quadratic equations?

Tip: Whenever solving polynomials with real coefficients, complex roots always appear in conjugate pairs!