Answer: (2x+3y) and (x-y)
Given:
[tex]2x^2+xy-3y^2[/tex]We can factor the given expression by grouping. We can rewrite the following expression:
[tex]\begin{gathered} ax^2+bxy-cy^2 \\ \Rightarrow(ax^2+uxy)+(vxy-cy^2) \end{gathered}[/tex]Given that:
a = 2
b = 1
c = -3
uv = ac
uv = 2(-3)
uv = -6
*Factor -6:
u = -2
v = 3
We now have:
[tex]\begin{gathered} 2x^{2}+xy-3y^{2} \\ \Rightarrow(2x^2-2xy)+(3xy-3y^2) \end{gathered}[/tex]We can now factor the expression easily:
[tex]\begin{gathered} \begin{equation*} (2x^2-2xy)+(3xy-3y^2) \end{equation*} \\ \Rightarrow2x(x^-y)+3y(x-y^) \\ \Rightarrow(2x+3y)(x-y) \end{gathered}[/tex]Therefore, its factors would be (2x+3y) and (x-y)