Determining if a quadratic polynomial is always positive Is there a quick and systematic method to find out if a quadratic polynomial is always positive or may have positive and negative or always negative for all values of its variables?
Say, for the quadratic inequality
$$3x^{2}+8xy+5xz+2yz+7y^{2}+2z^{2}>0$$
without drawing a graph to look at its shape, how can I find out if this form is always greater than zero or has negative results or it is always negative for all non-zero values of the variables?
I tried randomly substituting values into the variables but I could never be sure if I had covered all cases.
Thanks for any help.

This is what Sylvester's criterion is for. Write your quadratic as $v^T A v$ where $v$ is a vector of variables $(x_1\ x_2\ \cdots\ x_n)$ and $A$ is a matrix of constants. For example, in your case, you are interested in
$$\begin{pmatrix} x & y & z \end{pmatrix}
\begin{pmatrix} 3 & 4 & 5/2 \\ 4 & 7 & 1 \\ 5/2 & 1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}$$ 
Observe that the off diagonal entries are half the coefficients of the quadratic.
The standard terminology is that $A$ is "positive definite" if this quantity is positive for all nonzero $v$. Sylvester's criterion says that $A$ is positive definite if and only if the determinants of the top-left $k \times k$ submatrix are positive for $k=1$, $2$, ..., $n$. In our case, we need to test
$$\det \begin{pmatrix} 3 \end{pmatrix} =3 \quad \det \begin{pmatrix}3 & 4 \\ 4 & 7\end{pmatrix} = 5 \quad \det 
\begin{pmatrix} 3 & 4 & 5/2 \\ 4 & 7 & 1 \\ 5/2 & 1 & 2 \end{pmatrix} = -67/4.$$
Since the last quantity is negative, Sylvester's criterion tells us that this quadratic is NOT positive definite.
