Ashley...when you came across this problem...did you know what the solution would look like? I'm assuming that you knew what the problem was requiring you to do was find the values, that when multiplied together would give you the equation. x^2  5x + 6 The problem is asking you to find out what values, in this case, binomials, that when multipled together gives you the trinomial x^2  5x + 6 So what you're doing is dividing...in this case...finding the factors of the trinomial.. x^2  5x + 6 With that said, I'm assuming that you know your solution would be like: ( ?  ? )( ?  ? ) Correct? The problem is asking you to find the factors of the trinomial. Hint 1: The signs inside the trinomial tell you the signs inside the parenthesis will both be negative, like below. (  )(  ) Hint 2: the first values of the binomials will be x, because (x)(x) = x^s ( x  )( x  ) Hint 3: I'm assuming you know the FOIL methiod used in multiplying...and if you don't ask your instructor/teacher to help you with that, or you're welcome to drop me a line. Hint 4: The last value in x^2  5x + 6 is 6 at this point, the factors of 6 need to be identified...the values that when multiplied together the product is 6. The factors of 6 are 1, 1, 2, 2, 3, 3, 6, and 6...right, because 1 x 6 ... 2 x 3 ... 1 x 6... 2 x 3 correct? Well it is. Now here's a step that you can use Step 5: Filling in the parenthesis. Another way to find the other values in side the parenthesis. Look at each of the groups of factors. Take each group of factors and add the two values together: 1 + 6 = 7 and 2 + 3 = 5 2 + 3 = 5 1 + 6 = 7 Look at the sums of each of these...there is 5, 7, 7 and the other is 5. Notice that one group, the sum of the two factors is the same middle value in the trinomial This helps you determine with values you'll put in the parenthesis (2,3) so the solution would be ( x  3 )( x  2 ) or ( x  2 )( x  3) Here's another tool you can use to help you with these kinds of questions. See if it works with other similar kinds of problems. I hope this helps and the more you do these kinds of problems....then you'll be the MASTER
3/14/2013

Jon G.