how to factor? thanks :)

Question

9q²-18q+8 

Answer 1

1. When you have a trinomial of the form Ax² + Bx + C, you factor it by first finding two numbers that multiply to give you A*C, and that add up to B.
2. In this example, A = 9, B = -18 and C = 8.  So, you need to find two numbers that multiply to give you 9*8 = 72.  And they must add up to -18. 
3. After you make a complete list of integers that multiply to give you 72, you will find that the only numbers from this list that also add up to -18 are  -12 and -6.
4. Rewrite the original equation:  9q² - 18q + 8 =  9q² -12q -6q + 8  {Eqn 1}
5. Now group the first two terms of Eqn 1, i.e., 9q² -12q.   Factor out the 3q, and you get 3q(3q -4)
6. Similarly, group the last two terms of Eqn 1, i.e, -6q+8.  Factor out the 2, and you get -2(3q-4)
7. Now you combine terms:  (3q-2)(3q-4)  {Eqn 2}, and this is your answer
8. Always check your work when it's possible.  That's one of the nice things about math.  You can determine on your own if you got the right answer.
9. So multiply the two terms, using the FOIL Method:
1. First:  3q*3q = 9q²
2. Outside:  3q*-4 = -12q
3. Inside:  -2*3q = -6q
4. Last:  -2*-4 = 8
5. Combine like terms and you get 9q² -18q + 8.  This matches the original equation, and so you know that you factored correctly.

Answer 2

9q² - 18q + 8 = (3q - 4)(3q - 2)