The values of $k$ for which the equation
$x^2+2y^2-5z^2+2kyz+2zx+4xy=0$
represents a pair of plane passing Through origin,is
what i try
$x^2+2y^2-5z^2+2kyz+2zx+4xy=(ax+by+cz)(px+qy+rz)$
and camparing coefficients
but it is very tedious work
How do i solve it some short way Help me please
ADDED: conclusion without the matrices:
When $k^2 - 4 k - 8 = 0,$ we get factoring over the reals because $$ \color{magenta}{ (x+2y+z)^2 - \frac{1}{2} \left( 2y + (2-k) z \right)^2 } $$ is your quadratic form, and then we can factor as $$ T^2 - \frac{1}{2} U^2 = \left(T + \frac{U}{ \sqrt 2}\right) \left(T - \frac{U}{ \sqrt 2}\right) $$ when $T= x+2y+z$ and $U= 2y + (2-k) z.$ Note that $2-k$ simplifies when $k$ is one of the two roots of $k^2 - 4 k - 8 = (k-2)^2 - 12.$ Thus we have either $k = 2 + 2 \sqrt 3$ or $k = 2 - 2 \sqrt 3$
ORIGINAL::
we write a quadratic form as $X^T H X,$ where capital $X$ is the column vector with elements $x,y,z$ and $X^T = (x,y,z).$ Here, $H$ is the Hessian matrix or half of that, for convenience I'm taking half this time.
The form factors only if the determinant of $H$ is zero.
$$ H = \left( \begin{array}{ccc} 1&2&1 \\ 2&2&k \\ 1&k&-5 \\ \end{array} \right) $$
and $$ \det H = 8 + 4 k - k^2 $$
This becomes zero when $$ k = 2 \pm 2 \sqrt 3 $$
To make this concrete, I will display $Q^T D Q = H,$ where $D$ is diagonal and $\det Q = 1.$
$$ \left( \begin{array}{ccc} 1&0&0 \\ 2&1&0 \\ 1&\frac{2-k}{2}&1 \\ \end{array} \right) \left( \begin{array}{ccc} 1&0&0 \\ 0&-2&0 \\ 0&0&\frac{k^2 - 4k - 8}{2} \\ \end{array} \right) \left( \begin{array}{ccc} 1&2&1 \\ 0&1&\frac{2-k}{2} \\ 0&0&1 \\ \end{array} \right)= \left( \begin{array}{ccc} 1&2&1 \\ 2&2&k \\ 1&k&-5 \\ \end{array} \right) $$
When $k^2 - 4k - 8$ is nonzero, this expresses your quadratic form as a sum of three squares of linear terms, with coefficients.
When $k^2 - 4 k - 8 = 0,$ we get factoring over the reals because $$ (x+2y+z)^2 - 2 \left( y + \frac{2-k}{2} z \right)^2 $$ is your quadratic form, and then we can factor $$ V^2 - 2 W^2 = (V + W \sqrt 2)(V - W \sqrt 2) $$
We can assume that $a=p=1$, so we have $$x^2+2y^2-5z^2+2kyz+2zx+4xy=(x+by+cz)(x+qy+rz)$$ This should be true for all $y$, so it is true for $y=0$ also, and we get: $$x^2-5z^2+2zx=(x+cz)(x+rz)$$ So $cr=-5$ and $c+r=-2$. Also it is true for all $z$ and specialy for $z=0$: $$x^2+2y^2+4xy=(x+by)(x+qy)$$ so $bq=2$ and $b+q=4$...
$\begin{align} b + q =4\ bq = 2\ c + r = 2\ cq + br = 2k\ cr =-5 \end{align}$
The short way is to use a CAS like wolframalpha.
– Jan-Magnus Økland Nov 01 '19 at 14:52