Factorise $y^2 -3yz -10z^2$ How do I solve this question? I have looked at the problem several times. However, I cannot find a viable solution. I believe that it is a perfect square trinomial problem. 

$$y^2 - 3 y z - 10 z^2 = \begin{bmatrix} y\\ z\end{bmatrix}^T \begin{bmatrix} 1 & -\frac 32 + t\\ -\frac 32 - t & -10\end{bmatrix} \begin{bmatrix} y\\ z\end{bmatrix}$$
We want to find a $t$ that makes the matrix above rank-$1$. Computing the determinant,
$$\det \begin{bmatrix} 1 & -\frac 32 + t\\ -\frac 32 - t & -10\end{bmatrix} = t^2 - \left(\frac 72\right)^2$$
If $t = \frac 72$, then the matrix is singular and, thus, rank-$1$
$$\begin{bmatrix} 1 & 2\\ -5 & -10\end{bmatrix} = \begin{bmatrix} 1\\ -5\end{bmatrix} \begin{bmatrix} 1\\ 2\end{bmatrix}^T$$
Thus,
$$\begin{array}{rl} y^2 - 3 y z - 10 z^2 &= \begin{bmatrix} y\\ z\end{bmatrix}^T \begin{bmatrix} 1 & 2\\ -5 & -10\end{bmatrix} \begin{bmatrix} y\\ z\end{bmatrix}\\ &= \begin{bmatrix} y\\ z\end{bmatrix}^T \begin{bmatrix} 1\\ -5\end{bmatrix} \begin{bmatrix} 1\\ 2\end{bmatrix}^T \begin{bmatrix} y\\ z\end{bmatrix}\\ &= (y - 5 z) (y + 2 z)\end{array}$$

factoring polynomials quadratic-forms matrices rank-1-matrices
