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.

Answered by Tom L. Maths tutor

2963 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

(a) show that 3/10 + 2/15 = 13/30 (b) show that 2 5/8 ÷ 1 1/6 = 2 1/4


How do I sketch a quadratic function on graph paper?


How do you find the length of the diagonal of a rectangle of 5cm by 12cm?


Find the coordinates of the point where lines 3x+5=y and 6y+x=11 intersect


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences