Factorization of linear expressions

Factorisation is defined as the process of writing an algebraic expression as a product of its factors. It is the reverse process of expansion.

When we expand, we multiply out brackets: $3(a+2)=3a+6$. When we factorise, we do the reverse: $3a+6=3(a+2)$.

The correct description is writing an expression as a product of its factors, which matches the definition directly.

The question asks for the best description of factorisation. We need to select the option that correctly defines this process.

Expanding brackets to simplify an expression describes the reverse process — expansion, not factorisation. Eliminate this option.

Adding like terms is a simplification technique, not factorisation. This does not involve writing an expression as a product. Eliminate this option.

Substituting numbers into an algebraic expression is evaluation, which is completely different from factorisation. Eliminate this option.

The remaining option — writing an expression as a product of its factors — correctly defines factorisation.

$12k=2\times 2\times 3\times k$ $8k=2\times 2\times 2\times k$

The common prime factors of 12 and 8 are $2\times 2=4$. So the HCF of the numbers is $4$.

Both terms contain the variable $k$ (to the power of 1). So the HCF of the variable parts is $k$.

Overall HCF $=4\times k=4k$.

We need the largest factor that divides exactly into both $12k$ and $8k$, including both the numerical and variable parts.

$8k$ divides into $8k$ exactly, but $8k$ does not divide into $12k$ exactly ($12k÷8k=1.5$). So $8k$ is too large. Eliminate this option.

$4$ divides into both numbers (12 and 8), but it ignores the common variable $k$. Since both terms contain $k$, the HCF must include $k$. Eliminate $4$.

$2k$ is a common factor, but it is not the highest. Since $4k$ also divides both terms ($12k÷4k=3$ and $8k÷4k=2$), $2k$ is not the HCF. Eliminate $2k$.

$4k$ is the highest common factor — it includes the highest shared numerical factor (4) and the shared variable ($k$).