Since $2x+7$ is a factor, we should get a remainder of $0$ when we divide $6x^3+19x^2+cx+35$.
\[
\begin{array}{c|cccc}
\multicolumn{2}{r}{3x^2} & -x&+5  \\
\cline{2-5}
2x+7 & 6x^3&+19x^2&+cx&+35 \\
\multicolumn{2}{r}{-6x^3} & -21x^2  \\ 
\cline{2-3}
\multicolumn{2}{r}{0} & -2x^2 & +cx  \\
\multicolumn{2}{r}{} & +2x^2 & +7x \\ 
\cline{3-4}
\multicolumn{2}{r}{} & 0 & (c+7)x & + 35 \\ 
\multicolumn{2}{r}{} & & -10x & -35 \\ 
\cline{4-5}
\multicolumn{2}{r}{} & & (c+7-10)x & 0 \\ 
\end{array}
\]The remainder is $0$ if $c+7-10=0$, so $c=\boxed{3}$.
If $2x+7$ is a factor of $6x^3+19x^2+cx+35$, find $c$.
{'answer': '3', 'subject': 'Intermediate Algebra', 'level': 3, 'unique_id': 'train/intermediate_algebra/1912.json'}