Show algebraically that (4n-3)^2 - (2n+5)^2 is always a multiple of n-4

First we expand the brackets by squaring each side(4n-3)2 = (4n-3)(4n-3)= 16n2 - 24n + 9(2n+5)2 = (2n+5)(2n+5)= 4n2 + 20n + 25Remember the expression is (4n-3)2 - (2n+5)2 so we subtract the expanded second expression from the first16n2 - 24n + 9 - (4n2 + 20n + 25)= 16n2 - 24n + 9 - 4n2 -20n - 25= 12n2 - 44n - 4To simplify we can factorise by 44 (3n2 - 11n - 1)Then if we factorise the quadratic in the brackets we get4 (3n + 1)(n - 4)As the expression contains (n - 4), this means that (4n-3)2 - (2n+5)2 is always a multiple of (n - 4)

Answered by Ella B. Maths tutor

3033 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Expanding and simplifying, e.g. (x+4)(x-2)


A cuboid has sides such that the longest side is two units more than the shortest side, and the middle length side is one unit longer than the shortest side. The total surface area of the cuboid is 52 units². Calculate the length of the shortest side.


Solve: x^2 + y^2 = 25 y - 3x = 13


Solve the Simultaneous equations 4x - y = 8 and x + y = 12


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