Answer
$3(2r+1)(9r-10)$
Work Step by Step
Step 1. Let $u=3r-1$, we can rewrite the original expression as $6(3r-1)^2+(3r-1)-35=6u^2+u-35$
Step 2. Find factors of $6$ as $(2)(3)$, and find factors of $-35$ as $(5)(-7)$
Step 3. Find the sum of cross product of the factors as $(3)(5)+(2)(-7)=1$
Step 4. We have $6u^2+u-35=(3u-7)(2u+5)$
Step 5. Replace $u$ with $3r-1$, we have $(3u-7)(2u+5)=(3(3r-1)-7)(2(3r-1)+5)=(9r-10)(6r+3)=3(2r+1)(9r-10)$
Step 6. Thus $6(3r-1)^2+(3r-1)-35=3(2r+1)(9r-10)$
