9) (a -3)²
10) (3h -10)(2h +7)
There's an interesting method for factoring expressions of the second type. You write (6h + __)*(6h + __) where the numbers that go in the blanks are factors of 6*-70 (the product of first and last coefficients) that add to +1 (the h coefficient). Since those factors are nearly the opposite (their sum is 1), they are both near √420, which is near 20.5. So, the blanks get filled with -20 and +21 and you have
.. (6h -20)(6h +21)
This product is too large by a factor of 6, so we need to remove a factor of 6 from it. Neither binomial is divisible by 6, but the first is divisible by 2 and the second is divisible by 3. The result of dividing out those factors is shown above as your solution.