Factor 18abc2 - 12a2b2c completely.

Step-by-step solution

1. 1

   Take the GCF of the coefficients. $18=2\cdot 3^2$ and $12=2^2\cdot 3$, so their greatest common divisor is $2\cdot 3=6$. Taking only $2$ or only $3$ would leave a factorable remainder, so the factoring would not be complete.

2. 2

   Take the lowest power of each variable that appears in both terms. Compare exponent by exponent: $a^1$ versus $a^2$ gives $a$; $b^1$ versus $b^2$ gives $b$; $c^2$ versus $c^1$ gives $c$. The rule is always the minimum exponent, so the variable part of the GCF is $abc$.

3. 3

   Assemble the GCF.

   $\gcd=6abc$

   Every variable happens to appear in both terms here, which is why none of them is left out.

4. 4

   Divide each term by $6abc$.

   $\frac{18abc^2}{6abc}=3c,\qquad \frac{12a^2b^2c}{6abc}=2ab$

   so the bracket is $3c-2ab$, keeping the original minus sign between them.

5. 5

   Write the factorization and expand back.

   $18abc^2-12a^2b^2c=6abc(3c-2ab)$

   Expanding: $6abc\cdot 3c=18abc^2$ and $6abc\cdot 2ab=12a^2b^2c$ ✓. Numerically at $a=b=c=2$: the original is $18\cdot 16-12\cdot 32=288-384=-96$, and $6\cdot 8(6-8)=48(-2)=-96$ ✓. The bracket $3c-2ab$ has no common factor left, so the factoring is complete.