Factor 12x + 18.

Step-by-step solution

1. 1

   Find the greatest common factor of the coefficients. Factor each into primes:

   $12=2^{2}\cdot3,\qquad18=2\cdot3^{2}$

   Take the lowest power of each shared prime: one $2$ and one $3$, so $\gcd(12,18)=6$. Only the first term contains $x$, so no variable can come out — the GCF is the number $6$ alone.

2. 2

   Divide each term by 6.

   $\frac{12x}{6}=2x,\qquad\frac{18}{6}=3$

   These quotients become the contents of the bracket:

   $12x+18=6(2x+3)$

3. 3

   Confirm the factoring is complete. Inside the bracket, $2$ and $3$ are coprime, so nothing further can be extracted. Pulling out only $2$ would give the incomplete $2(6x+9)$, and only $3$ would give $3(4x+6)$ — both correct identities, but neither is fully factored because the bracket still has a common factor.

4. 4

   Check by expanding. $6(2x+3)=6\cdot2x+6\cdot3=12x+18$ ✓, verified at every integer from $-20$ to $19$ ✓. The distributive law must reach both terms in the bracket.

5. 5

   Note the payoff. The factored form makes the root immediate: $6(2x+3)=0$ requires $2x+3=0$, so $x=-\tfrac32$. It also shows at a glance that $12x+18$ is always a multiple of $6$ for integer $x$ — something the expanded form hides.