Question 139591
There is a slight flaw in your terminology.  ALL 2nd degree trinomials can be expressed as the product of two factors, {{{x-r}}} and {{{x-s}}}.  The thing is, r and s aren't always integers or even rational numbers.  I suspect your instructor meant to say "If the polynomial is not factorable <i>over the integers (or rationals),</i> write prime."


With this ungodly horror, {{{2(x-y)^2-9(x-y)z-5z^2}}}, recognize that the independent variable is {{{x-y}}}, so let's clean it up a bit by saying {{{u=x-y}}}.  That let's us rewrite the expression: {{{2u^2-9z*u-5z^2}}}.  Now we can see that the pattern {{{ax^2+bx+c}}} means that {{{a=2}}}, {{{b=-9z}}}, and {{{c=-5z^2}}}


Now we can use the fact that {{{r}}} and {{{s}}} are roots of {{{ax^2+by+c=0}}} if and only if {{{x-r}}} and {{{x-s}}} are factors of {{{ax^2+by+c}}}.


So let's find the roots of {{{2u^2-9z*u-5z^2=0}}} using the quadratic formula:
{{{u = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


{{{u = (-(-9z)+- sqrt( (-9z)^2-4*2*(-5z^2) ))/(2*2) }}}


{{{u = (9z+- sqrt( 81z^2+40z^2 ))/(4) }}}


{{{u = (9z+- sqrt(121z^2 ))/(4) }}}


{{{u = (9z+11z)/(4) }}} or {{{u = (9z-11z)/(4) }}}


{{{u = (20z)/(4) }}} or {{{u = (-2z)/(4) }}}


{{{u = 5z }}} or {{{u = -z/2 }}}


That means that either {{{u-5z=0}}} or {{{u + z/2 =0}}} (which is equivalent to {{{2u+z=0}}}.


Hence our factors so far are {{{(u-5z)(2u+z)=0}}}, but remember that {{{u=x-y}}}, so:


{{{((x-y)-5z)(2(x-y)+z)}}} are your two factors.