# How is the resultant $R(z) =\text{res}_θ(1 - \frac{z}{x}, θ) = 1 - \frac{z}{x}$ calculated?

The resultant of two polynomials is defined as the determinant of the Sylvester matrix. If the polynomials are of degree $n$ and $m$, than the Sylvester matrix will be of dimension

$(m+n)\times (m+n)$.

In the book Algorithms for Computer Algebra on page 534 (example 12.6) the following resultant is calculated:

$R(z) =\text{res}_θ(1 - \frac{z}{x}, θ) = 1 - \frac{z}{x}$

How is this resultant calculated?

• I'm not quite sure what to make of your notation. The expression for $R(z)$ seems to deal with polynomials in more variables, while the Sylvester-version only deals with one variable. What am I missing?
– HSN
Commented May 18, 2013 at 2:06
• @HSN, that's exactly what I couldn't figure out ))) Vinicius M.'s answer clarifyes that Commented May 18, 2013 at 16:22

I guess the point is to consider $\theta$ as the variable. So the resultant in this case is a $1\times 1$-matrix (since the degree of $1-\frac{z}{x}$ as a polynomial in the indeterminate $\theta$ is $0$ and the degree of $\theta$ is $1$), precisely $[1-\frac{z}{x}]$.