Cuando factorizamos un polinomio de esta forma, estamos básicamente revirtiendo este proceso de multiplicar expresiones lineales:

\qquad \begin{eqnarray} (x + ay)(x + by)&=&xx &+& xby + ayx &+& ayby \\ \\ &=& x^2 &+& \green{(a+b)}xy &+& \blue{ab}y^2 \\ &\hphantom{=}& \hphantom{x^2} &\hphantom{+}& \hphantom{\green{LINEAR}xy} &\hphantom{+}& \hphantom{\blue{CONSTANT}y^2} \end{eqnarray}

\qquad \begin{eqnarray} \hphantom{(x + ay)(x + by)}&\hphantom{=}&\hphantom{xx} &\hphantom{+}& \hphantom{xby + ayx} &\hphantom{+}&\hphantom{ayby} \\ &\hphantom{=}& \hphantom{x^2} &\hphantom{+}&\hphantom{\green{(a+b)}xy}&\hphantom{+}&\hphantom{\blue{ab}y^2} \\ &=& x^2 &+& \green{LINEAR}xy &+& \blue{CONSTANT}y^2 \end{eqnarray}

The coefficient on the xy term is LINEAR and the coefficient on the y^2 term is CONSTANT, so to reverse the steps above, we need to find two numbers that add up to LINEAR and multiply to CONSTANT.

Puedes empezar tratando de adivinar qué factores de CONSTANT sumados dan LINEAR. En otras palabras, necesitas encontrar los valores de a y b que satisfacen las siguientes condiciones:

\qquad \pink{a} + \pink{b} = \green{LINEAR}

\qquad \pink{a} \times \pink{b} = \blue{CONSTANT}

If you're stuck, try listing out every single factor of CONSTANT and its opposite as a in these equations, and see if it gives a value for b that validates both conditions. For example, since abs(A) is a factor of CONSTANT, try substituting abs(A) for a as well as -abs(A).

Los números -A y -B satisfacen las dos condiciones:

\qquad \pink{-A} + \pink{-B} = \green{LINEAR}

\qquad \pink{-A} \times \pink{-B} = \blue{CONSTANT}