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<math>a(a-3b)+4c(a-3b)</math><br /><br />
'''Step 3''' If the 'leftovers match' factor them out.<br />
Since there is a (a-3b) in each term, factor out (a-3b) from each term.
<math>a\mathbf{(a-3b)}+4c\mathbf{(a-3b)}=(a-3b)(a+4c)</math><br /><br />
Therefore <math>a^2-3ab+4ac-12bc=(a-3b)(a+4c)</math><br /><br /><br />
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