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. 
 A: As Andre pointed, you can guess the factors by inspection.
Alternately, we can factor $z^2$ and then you get
$$z^2[(\frac{y}{z})^2-3 \frac{y}{z}-10] \,.$$
Denoting $\frac{y}{z}=w$ the problem reduces to factoring 
$$w^2-3w-10$$
This can be factored as $(w-a)(w-b)$ where $a,b$ are the two roots of $w^2-3w-10=0$.
Calculate the roots, and remultiply by $z^2$ to get the factorization.
A: $$y^2 - 3 y z - 10 z^2 = \begin{bmatrix} y\\ z\end{bmatrix}^T \begin{bmatrix} 1 & -\frac 32 + t\\ -\frac 32 - t & -10\end{bmatrix} \begin{bmatrix} y\\ z\end{bmatrix}$$
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
A: Completing the square,
$$y^2-3yz-10z^2=\frac{4y^2-12yz-40z^2}4=\frac{(2y)^2-2\cdot2y\cdot3z+(3z)^2-49z^2}4$$
$$=\frac{(2y-3z)^2-(7z)^2}4=\frac{\{(2y-3z)-7z\}\{(2y-3z)+7z\}}4=\cdots$$
A: I made up a method with no guesswork at all, Factoring Quadratics: Asterisk Method 
and my earlier answer at How to factor the quadratic polynomial $2x^2-5xy-y^2$? 
We require that $b^2 - 4ac$ be a perfect square, and (positive) $\delta = \sqrt {b^2 - 4ac},$ otherwise factoring is impossible anyway. Then: we may simply take $$ a_1 = \gcd \left( a, 
  \frac{(b + \delta)}{2}  \right) ; \; \; \; \; \; a_2 = \frac{a}{a_1}  $$ without paying attention to any prime factorizations and
$$  a x^2 + b x y + c y^2 =  \; \left(a_2x+ \left( \frac{b + \delta}{2a_1} \right) y \right) \; \; \left(a_1x+ \left( \frac{b - \delta}{2a_2} \right) y \right) \;    $$ in integers.
You have $a=1, b= -3, c=-10, b^2 - 4 a c = 49, \delta = 7.$ So $a_1 = 1, a_2 = 1.$ Then $(b+\delta)/ (2 a_1) = 4/2 = 2,$ and $(b-\delta)/ (2 a_2) = -10/2 = -5,$ so $x^2 - 3 x y - 10 y^2 = (x + 2 y)(x - 5 y).$
A: $y^2-3yz-10z^2=y^2-5yz+2yz-10z^2=y(y-5z)+2z(y-5z)=(y-5z)(y+2z)$
