Are there any quick ways to find two numbers that can both be multiplied and added to reach certain numbers? For example: If I wanted to find a number that multiplied to 14 and added to 9, I would use 7 and 2 because 7*2=14 and 7+2=9. You use this a lot when factoring by grouping, and I was wondering if there were any tips to finding the two numbers faster, especially when dealing with larger numbers that have more factors.
Are there any quick ways to factor by grouping when dealing with large numbers?
No matter how big the number is, you should be able to list its factors: For example, if the number is 36, then its factors are 9 x 4, 6 x 6, 12 x 3. So either the 9 & 4 or 6 & 6 or 12 & 3 will go into each of your parenthesis when grouping. Before trying them look to see what the value of your X is in the middle of the equation and which of those two numbers when added or subtracted will give you the final result.
Example: x^2 + 9x -36. By seeing the 36 at the end, we know we must use a set of the factors listed above. But which set from above will give you +9x in the middle, it must be a +12 and a -3. Hope this helps :)
Use a multiplication chart and then you can easily see if the 2 numbers multiply to the given number and then add them up and see if that works. It works well if you do not have your multiplication tables memorized. If you have your multiplication tables memorized then factoring should be fairly easy for you! :) Good Luck
Factoring by grouping usually looks something like this:
x^2-2x-1+2xy. And you have to find the best way to factor the math sentence?
If u are using numbers; prime factoring is the best way or using the GCF(greatest common factor) would give you your answer to finding the easiest common factor that can be used. Reply if you want step by steps to a problem? :)
When dealing with large numbers, I suggest breaking them down in the prime factors. For example, 180 would be 2*2*3*3*5. This is helpful because let's say you are trying to find 2 factors of 180 that add up 27. Since you have the prime factors of 180 listed, you just need to try different combinations until you come up with the answer. In this case, I tried (2*2*3) and (3*5), which is 12 and 15.