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)