The common factor of 2a2b, 4ab2 and 6ac2 is:

The correct answer is Option (2) → 2a

A common factor of algebraic terms must divide every one of the terms. Writing each term as a product of its factors shows what all three share.

• \(2a^2b=2\times a\times a\times b\)

• \(4ab^2=2\times2\times a\times b\times b\)

• \(6ac^2=2\times3\times a\times c\times c\)

• Every term has one 2 and one \(a\), so the common factor is \(2a\)

Why the other options are incorrect

• Option (1) 2ab — Incorrect. \(6ac^2\) has no factor \(b\), so 2ab does not divide it.

• Option (3) 2abc — Incorrect. Neither \(2a^2b\) nor \(4ab^2\) has a factor \(c\).

• Option (4) 2b — Incorrect. \(6ac^2\) has no factor \(b\).