In your first example, imagine you let y = a, and you found out that the result was indeed zero! Then you could say that is a factor of the polynomial. You would then divide the original polynomial by using long division/synthetic division. This will enable you to write the expression in the form:
, where is a quadratic that can be solved using the quadratic equation/factorisation!
In question 1, guessing y = 2 gives you a root, hence is a factor. If you divide , by , you get . Hence:
You can factorise that quadratic easily to find the other roots!
For the last 3 problems you have rearrange them to get 0 on the RHS, and you also need the expland the LHS of 3).
2.) Which is a cubic, so apply the rules I used before! Hint: x = 2 is a root.
3.) Which is a quadratic, easily solveable using factorisation.
4.) . Another cubic, so apply the rules I laid out for question 1! Hint: x = 4 is a root.