Question 464766
We can let y = 5a-3 and we get a simpler-looking quadratic:


*[tex \LARGE (5a-3)^2 - 6(5a-3) + 9 = y^2 - 6y + 9]



The quadratic factors:


*[tex \LARGE y^2 - 6y + 9 = (y-3)(y-3)]


We want to express in terms of a, not y, so we substitute:


*[tex \LARGE ((5a-3)-3)((5a-3)-3) = (5a-6)(5a-6)]