Factorise fully 6xyz + 24x^2yz + 18xy^3z^2

So we need to find common factors for each term: 6xyz, 24x²yz and 18xy³z². Let's start with the numbers. Do the numbers 6, 24 and 18 have a common factor? Yes, 6 . If we take 6 out of the term we get: 6(xyz + 4x²yz + 3xy³z²). Now we need to factorise out the letters. Each term has a common factor of xyz, but nothing more. For example, as 4x²yz has an x² in it but xyz does not, so we cannot take out x². This gives us then: 6xyz(1 + 4x + 3y²z) which is fully factorised. To check your answer you can try multiplying back out the bracket and seeing if you end back with 6xyz + 24x²yz + 18xy³z².