## Intermediate Algebra (6th Edition)

$(3x+2)(3x+1)$
$(3x-1)^{2}+5(3x-1)+6$ By substituting $3x-1 = t$, we get, $t^{2}+5t+6$ In this trinomial $t^{2}+5t+6$, $a=1$, $b=5$ and $c=6$. Splitting the middle term $b$ into two numbers whose product is $6$ $(a \times c)$ and whose sum is $(b)$ $5$. The numbers are $3$ and $2$. $=t^{2}+3t+2t+6$ Factor out by grouping. $=(t^{2}+3t)+(2t+6)$ $=(t(t+3)+2(t+3))$ $=(t+3)(t+2)$ Substituting $t$ value, $=(3x-1+3)(3x-1+2)$ $=(3x+2)(3x+1)$