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

3034 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve( 3x−2)/4 −(2x+5)/3= (1−x )/6


Solve the simultaneous equations 2x - 3y = 24 (1) ; 6x + 2y = -5 (2)


Simplyfy, ((x-2)(x^2+5x+6)-(2x^2+10x+12))/(x^2+x-2)


Solve the simultaneous equations, 3x + 2y = 4 (1) 4x + 5y = 17 (2)


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