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)
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