Answer
$(x+3)$ and $(x-2)$,
$(x+3)$ and $(x+1)$,
$(x+3)(x-2)(x+1)$
Work Step by Step
Factoring trinomials of type $x^{2}+bx+c,$
we search for two factors of c whose sum is b...
$x^{2}+x-6=$
... we find $-2$ and $3$ (product =-6, sum =1)
$=(x+3)(x-2)$
$x^{2}+4x+3=$
... we find $+3$ and $+1$ (product =$3$, sum =$4$)
$=(x+3)(x+1)$
The LCD is formed so
- we list any common factors in both denominators : $(x+3)$
- multiply with all "leftover" factors.
$LCD=(x+3)\times(x-2)(x+1)$
Fill the blanks with
$(x+3)$ and $(x-2)$,
$(x+3)$ and $(x+1)$,
$(x+3)(x-2)(x+1)$
