#### Answer

Fill the blank with $3x+4.$

#### Work Step by Step

$ax^{2}+bx+c=(F_{1}x+L_{1})(F_{2}x+L_{2})$
FOIL: (F)irst, (L)ast...Outers: $F_{1}$and $L_{2},$ Inners=$ L_{1}$ and $F_{2}$
The product of Firsts is $ax^{2}$
The product of Lasts is $c$
The sum of Outside product and Inside product is $b$
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The product of Firsts is $6x^{2} $
... for this to work, $F_{2}$ must be $ 3x$
The product of Lasts is $+12$
... for this to work, the last term in the second parentheses should be $+4.$
Proposed factorization: $(2x+3)(3x+4)$
The sum of Outside product and Inside product is $17$
$2(4)+(3)(3)= 17\qquad ... (OK)$
Fill the blank with $3x+4.$