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There is no common factor that can be factored out from **all** four terms. However, we note that $2x_{2}$ and $2x$ share the common factor $2x.$ Further, we can also see that $3x$ and $3$ share the common factor $3.$ Begin by factoring these out.
Now we only have two terms and we can see that both contain $(x+1).$ It means that it can factored out.

$2x_{2}+2x+3x+3$

SplitIntoFactorsSplit into factors

$2x⋅x+2x⋅1+3⋅x+3⋅1$

FactorOutFactor out $2x$

$2x(x+1)+3⋅x+3⋅1$

FactorOutFactor out $3$

$2x(x+1)+3(x+1)$