Factorization using identities

The two terms are $6x$ and $18$. The coefficients are 6 and 18. $\text{HCF}(6,18)=6$.

The term $6x$ has variable $x$, but $18$ is a constant with no variable. So there is no shared variable to include in the HCF.

$6x÷6=x18÷6=3$ So the expression inside the brackets is $(x+3)$.

$6x+18=6(x+3)$ Check: $6(x+3)=6x+18$ ✓. This is fully factorised because $(x+3)$ cannot be simplified further.

We need the fully factorised form, meaning the HCF must be taken out completely — nothing inside the bracket should share a common factor.

Inside the bracket, $2x+6$ still has a common factor of 2: $2x+6=2(x+3)$. So $3(2x+6)$ is only partially factorised.

Inside the bracket, $3x+9$ still has a common factor of 3: $3x+9=3(x+3)$. So $2(3x+9)$ is also only partially factorised.

Expanding gives $6x+108$, which does not equal $6x+18$. This option is simply incorrect.

$6(x+3)$ uses the full HCF of 6, the bracket $(x+3)$ has no further common factor, and expanding gives $6x+18$ ✓.

The coefficients are 12 and 8. $\text{HCF}(12,8)=4$.

The variable parts are $x^2$ and $x^1$. Take the lowest power: $x^1=x$.

$\text{HCF}=4\times x=4x$

$12x^2÷4x=3x$ and $8x÷4x=2$, both of which are integers/simple terms. So $4x$ is correct.

The HCF must divide both $12x^2$ and $8x$ exactly, and it must be the largest such factor.

$8x÷4x^2=\frac{2}{x}$, which is not a polynomial term. So $4x^2$ does not divide $8x$ exactly.

$12x^2÷8x=\frac{3x}{2}$, which is not an integer coefficient. So $8x$ does not divide $12x^2$ exactly.

$2x$ does divide both terms, but it is not the highest common factor. Since $4x$ also divides both, $2x$ is too small.

$4x$ divides both $12x^2$ (giving $3x$) and $8x$ (giving $2$) exactly, and no larger factor does. So $4x$ is the HCF.