Factorise $x^{2}+4 y^{2}+25 z^{2}+4 x y-20 y z-10 z x$.

(N/A) The given expression is $x^{2}+4 y^{2}+25 z^{2}+4 x y-20 y z-10 z x$.
We can write this in the form of the algebraic identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$.
Here,$a = x$,$b = 2y$,and $c = -5z$.
Substituting these values into the identity:
$x^{2} + (2y)^{2} + (-5z)^{2} + 2(x)(2y) + 2(2y)(-5z) + 2(-5z)(x)$
$= x^{2} + 4y^{2} + 25z^{2} + 4xy - 20yz - 10zx$.
Thus,the expression is equivalent to $(x+2y-5z)^{2}$.
Therefore,the factors are $(x+2y-5z)(x+2y-5z)$.