Fatorar: $\phantom{X}x^{{}^{\Large 2}}\,+\,2y^{{}^{\Large 2}}\,+\,3xy\,+\,x\,+\,y\phantom{X}$
resposta:
Resolução:
$\,x^{{}^{\Large 2}}\,+\,2y^{{}^{\Large 2}}\,+\,3xy\,+\,x\,+\,y\,=$ $\,x^{{}^{\Large 2}}\,+\,2y^{{}^{\Large 2}}\,+\,xy\,+\,2xy\,+\,x\,+\,y\,=$ $\,\,x^{{}^{\Large 2}}\,+\,xy\,+\,2y^{{}^{\Large 2}}\,+\,2xy\,+\,x\,+\,y\,=$ $\,x(x\,+\,y)\,+\,2y\centerdot(x\,+\,y)\,+\,1\centerdot(x\,+\,y)\,=$ (x + y)(x + 2y + 1)×