Let's rewrite the equation in operational form
"F(D,D')=aD^2+2bDD'+cD'^2=0"
An equation is irreductable if it can't be written as a product of linear factors, hence
"F(D;D')=(a_1D+b_1D'+c_1)(a_2D+b_2D'+c_2)=a_1a_2D^2+"
"+(a_1b_2+a_2b_1)DD'+b_1b_2D'^2+(a_1c_2+c_1a_2)D+(b_1c_2+c_1b_2)D'+c_1c_2"
Using the method of undetermined coefficients, we assume
"a_1a_2=a; b_1b_2=c; a_1b_2+a_2b_1=2b;"
"b_1c_2+c_1b_2=a_1c_2+c_1a_2=c_1c_2=0"
Hence "c_1=c_2=0"
If "b^2-ac=0," "a_1=a_2=\\alpha; b_1=b_2=\\beta" ,therefore "F(D,D')=(\\alpha D+\\beta D')^2" which is irreducible
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