Fatorar: $\phantom{X}x^{{}^{\Large 2}}\,-\,5x\,+\,6\phantom{X}$
resposta:
Resolução:
$\,x^{{}^{\Large 2}}\,-\,5x\,+\,6\,=\,\rlap{ \underbrace{\phantom{x^{{}^{\Large 2}}\,-\,2x\,}}_{\small x(x\,-\,2)} }x^{{}^{\Large 2}}\,-\,2x\, - \,\underbrace{3x\,+\,6}_{\small -3(x\,-\,2)} \,= $ $\,x(x\,-\,2)\,-\,3\centerdot(x\,-\,2)\,=\,$ (x - 3)(x - 2)×