Hello Samira,
The polynomial 2(y-3)2 - 5(y-3) + 3 is quadratic in form. This means that if you make an appropriate substitution, the resulting polynomial will be quadratic (second-degree) in the substitution variable. In this case, if you let
U = y-3, then we may write
(Eq. 1) 2(y-3)2 - 5(y-3) + 3 = 2U2 - 5U + 3.
Using trial and error (or some other method you may have learned), we can factor 2U2 - 5U + 3 and obtain
2U2 - 5U + 3 = (2U - 3)(U - 1).
To complete the factorization, use this result in Eq. 1, back-substitute U = x - 3, and simplify:
2(y-3)2 - 5(y-3) + 3 = (2U - 3)(U - 1)
= [2(y - 3) - 3][(y - 3) - 1]
= (2y- 6 - 3)(y - 3 - 1)
= (2y - 9)(y - 4).
Hope that helps. Let me know if you have any questions.
William
