How do you factor $6 a b c^{2} - 14 a^{2} b c + 4 a b^{2} c$ $6abc^(2) - 14a^(2)bc + 4ab^(2)c$ ?

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Answer 1 · 2016-06-30T16:33:50

$2 a b c (3 c - 7 a + 2 b)$ $2abc(3c-7a+2b)$

Always look for a common factor first.
There is a common factor of $2 a b c$ $2abc$ which can be divided out of each term.

$6 a b c^{2} - 14 a^{2} b c + 4 a b^{2} c$ $6abc^(2) - 14a^(2)bc + 4ab^(2)c$

= $2 a b c (3 c - 7 a + 2 b)$ $2abc(3c-7a+2b)$

How do you factor 6abc2−14a2bc+4ab2c6abc^(2) - 14a^(2)bc + 4ab^(2)c?