How to know if a non-homogeneous quadratic function with two variables can be factorized and how to factorize it?

Question

I'm talking about functions like: $$p^2-3pq+2q^2-p+3q-2$$ It is fairly easy to factorize homogeneous ones but I don't understand how to deal with non-homogeneous ones. Help?

Answer 1

You can set up the factorization as $$(p+Aq+B)(p+Cq+D)$$ and arrive fairly easily, by comparing coefficients, with equations like $$AC=2$$$$A+C=-3$$ so symmetrically $A=-1,C=-2$ and similarly obtain the values of $B$ and $D$

Answer 2

$$p^2-3pq+2q^2-p+3q-2=p^2-(3q+1)p+2q^2+3q-2=$$ $$=p^2-(3q+1)p+(2q-1)(q+2)=p^2-(2q-1+q+2)p+(2q-1)(q+2)=$$ $$=(p-(2q-1))(p-(q+2))=(p-2q+1)(p-q-2).$$

Answer 3

Start as David Quinn suggested: $$p^2-3pq+2q^2-p+3q-2=(p+Aq+B)(p+Cq+D).$$ Then plug $q=0$ and $p=0$: $$p^2-p-2=(p+B)(p+D) \Rightarrow B=2; \ D=-1;$$ $$2q^2+3q-2=(Aq+2)(Cq-1) \Rightarrow A=1; C=2.$$

Answer 4

$p^2-(3q+1)p+2q^2+3q-2$ is a quadratic in $\,p\,$ with discriminant:

$$ \Delta = (3q+1)^2-4(2q^2+3q-2)=q^2 - 6 q + 9=(q-3)^2 $$

Therefore the roots in $\displaystyle\,p\,$ are $\,p_{1,2}=\dfrac{3q+1\pm \sqrt{\Delta}}{2}=\dfrac{3q+1\pm (q-3)}{2}=\begin{cases}2q-1 \\q+2\end{cases}\,$, and the quadratic factors as $\,(p-p_1)(p-p_2)\,$.