Correlation between polynomial equations and matrix determinants Expanding
$p(x)=(ax-b)(cx+d)$
we get
$acx^2+(ad-bc)x-bd$.
Notice the determinant of the matrix
$\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
$
is $ad-bc$ exactly like the constant of $x$ in the polynomial expansion.
So I searched links between the equation and the matrix determinant. I only found the solution of the equation $p(x)=0$ to be $x \in \{b/a, -d/c\}$. Then I found numerous other things, but none of them were simple enough to be considered worthy remembering.
My question: What are the correlations between the polynomials and determinants of the matrices with entries being the coefficients of the polynomial?
 A: It is not entirely clear to me what you are asking, but you might be interested in knowing that any monic polynomial (say with real coefficients, but any ring will do) occurs as the characteristic polynomial of some matrix with coefficients in $\mathbb R$ (or the same field the coefficients of the polynomial are coming from). It's a nice exercise to find the matrix (hint: it's more straightforward than one might initially think).
A: A very important link between polynomials and determinants is this: if $A$ is a square matrix, $I$ is the identity matrix, and $x$ is an indeterminate, then $\det A$ is the constant term of the polynomial $\det(A-xI)$. 
A: Conceptually, one nice way to "explain" this is as follows. We know from high-school that for a polynomial $\rm\: x^n - \color{#C00}{c_{n-1}}x^{n-1}\!+\,\cdots+c_0,\:$ the coefficient $\rm\,\color{#C00}{c_{n-1}}\,$ equals the sum of the roots. Hence $\rm\: \ell(f)\, :=\, \color{#C00}{c_{n-1}}$ satisfies a $\rm\color{blue}{logarithmic}$ law $\rm\ \ell(fg)\, =\, \ell(f)+\ell(g)\ $ (as is also easily verified directly). Your observation may be interpreted as a variation on this logarithmic property of root sums.
Define $\rm\,\ f\,\left[\begin{array}{cc}\rm a&\rm b\\ \rm c&\rm d\end{array}\right]\, :=\ \ell((ax\!-\!b)(cx\!+\!d))\, :=\, $ coefficient of $\rm\,x\,$ (the "generic" value of $\ell)$  


*

*$\rm\,f\,$ is multilinear: $\, $ True, essentially, by the above $\rm\color{blue}{logarithmic}$ property of $\:\ell$.

*$\rm\,f\,$ is alternating:  $\rm\, \ f\,\left[\begin{array}{cc}\rm a&\rm b\\ \rm a&\rm b\end{array}\right] =\ \ell((ax\!-\!b)(ax\!+\!b))\, =\, \ell(a^2x^2\!+\color{#C00}0\cdot x-\!b^2)\, =\, \color{#C00}0$

*$\rm\,f\,$ is normalized: $\rm\ f(I) = 1\:$ by $\rm\: f\,\left[\begin{array}{cc}\rm 1&\rm 0\\ \rm 0&\rm 1\end{array}\right] =\ \ell(1\!\cdot\! x\!-\!0)(0\!\cdot\! x\!+\!1))\, =\, \ell(\color{#C00}1\cdot x) = \color{#C00}1$
However, as is well-known, $\rm\:det\,$ (determinant) is the unique solution of the above equations. Therefore, by uniqueness, we deduce that $\rm\ f = det.$
