Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Factoring leading coefficients. We're going to take trinomials and learn a method to factor. There are many, many, many different methods. This is one that I find particularly easy and useful. So we're going to multiply the coefficient on the squared term or the leading term and multiply the constant at the end. For right now, I'm not going to worry about the signs of the pluses and minuses. I'm just going to pay attention to the 6 * 35. Well, 6 * 35 is 210, so I'm going to multiply. I'm going to list all of the two numbers that multiply to give me 210. This is a great place to practice your division and factoring. So one and 2:10 210 is even. So I know that A2 can come out. If I pull out A22 goes into that are leading to one time 2 doesn't go into one at all, but two then goes into the 10 because we didn't use the one a minute ago five times. Now I know that I can't take another two out because one O 5 is odd. So I know that 481632 etcetera won't go in, but what about 3? The divisibility test for three says add up all the digits and if all the digits is a multiple of 3 then the original number will be. So 2 + 1 + 0 is 3. So I know that three goes into 210. So 3 and 74 we knew couldn't because this one O 5 was not even 5 is going to because the number ends in A0 or A5. SO55 goes into 21 four times with one leftover. So now think about 5 going into 10 two times. Not using my calculator for any of this. I'm practicing my skills. So six, Well, if two goes in and three goes in, the fact that two and three have nothing in common tells me that six has to go in also. Now one of the really cool tricks is if I think about 3 * 70, I could take two out of the 70 and make this into 2 * 35 S the three times the two would give me 6 and that would leave me a 35 over here. Well, does 7 go in? If seven goes into any of these numbers that we can see easily, we know it goes into the original. Well, 70 here is 7 * 10, so if I pull out the seven, I'd have a 10 left here and a three there. So 7 * 30 is 210. We knew 8 couldn't because back here we didn't have another two to pull out of 105 nine. The divisibility test for 9 is exactly the same for three, but it has to be a multiple of nine. 2 + 1 + 0 is 3. 9 doesn't divide evenly into 3, so 9's not a factor. 10 if it ends in a 0, so 10 and 21 and then 11 won't. 12 won't because 4 didn't. 13 won't, but 14 does because two and seven both did, and two and seven don't have any common factors amongst themselves. So 14 if I thought about taking a two out of this 30, so 7 times that 2 is 14. Two out of 30 gives me 15. I've now listed every single possible factor of 2, numbers that multiply to give me 210. Now I'm going to pay attention to the signs. One of these was positive and the other negative. So I need to have the multiplied to give me a negative in the end. But I want the sum to be a positive and I get that from this middle term. So if I want them to multiply to give me a negative but add to give me a positive, I know that the smaller number is going to be a negative. So I would come through here and I would put a negative on each of the smaller numbers. So now we're going to find the sum. So this would be two O 9, one O3 6738 Nope, 37 sorry 2923 eleven and one. So when I look at this middle term, I need the one that has a one. So the numbers I'm going to use are -14 and 15. Now that may have all seemed very familiar. This next part maybe you haven't seen before. I'm going to take and put X * X in the front location, and I'm going to put the -14 and the +15 in the numerators divided by the leading coefficient of the six right there. Now I'm going to reduce those. So 14-6 is 7 thirds, 2 divides out of each 15-6. A3 comes out of each is five halves. Once I've reduced them, I'm going to pull the denominator out in front and that is going to give me my correct factorization 3X times. 2X was the six X squared. The outer term of 15X and the inner term of -14 X would add to give me One X. And then the last terms of -7 and +5 do indeed give me -35. So I've given us several to practice on that would use the same column of numbers. So the 6 * -35 we'd get 2:10. We'd list them all like we did a minute ago, but now we want them to add to give me this 11 in the middle +11. So when we look over here, if I want +11, I'm going to use -10 and 21. So if I do X - 10 / 6 and X + 21 / 6, reduce those so X -, 5 thirds and X + 7 halves, so three X - 5 and two X + 7, and I'm done factoring. No guessing and checking here at all, no making squares. Just a nice crisp straightforward method to get the answer. So this next one, I need it to be a -29. If I need it to be a negative, that actually tells me that the bigger number is going to be the negative, so that when I take the sum, a bigger one that's negative with a smaller than one that's positive would give me all these being negatives. And if I want it to be a -29, I'm going to use +6 and -35 South +6 negative 30, five X + 6 / 6 X -35 / 6. When I reduce, I get X + 135 and 6 didn't have anything in common. So my final factorization, X + 1 and six X -, 35, we can foil that out to check. So six X * X is six X ^2 -35 X +6 X and -35. And we can see that really does give us what we wanted. Let's do another example so that you we can see it all the way through again. So this time we're going to multiply the 8 and the fifteen. 8 * 15 is 120, so 8 * 15120. So we have one and 122 and 60. Now 60 is Even, so we know that another two could come out. If I pull A2 out of here, I would get four with 30 left because 60 is 2 * 31 of the twos times this 2. Now 30 is still even, so I can pull another two out and get 8 * 1515 is not even. So the only multiples of 2 powers of two that can come out are two, four, and eight. Going back, let's see can 3 come in and three evenly divides in because 1 + 2 + 0 is a multiple of 3. So 3 * 45 is going to work because 5 has the number ending in A0 or A55 goes into 12 twice with two leftover and then five goes into 24 times. Will 6 work? Two and three both worked and two and three have nothing in common so 6 has to work. I can come up to this 3 and 40 and think about if I divided 2 out of the 43 * 2 gave me the six so that would leave me 27 is not going to work. 8 we realized did nines not because 1 + 2 + 0 is not a multiple of 9-10 will work because it ended in a 0 and then 11 doesn't work. And now we're back to all the bigger numbers. We don't have to test them because once we got the two in the middle that worked, all the rest were there already there. They were paired with something. So now we look at this middle sign and it's positive. So I'm going to have the sum adding to give us a positive, and in this case the two numbers on the outside were both positive, so we needed them to multiply to give us a positive and add to give us a positive. So 12162433429262322. Now let me emphasize once again, I'm not using a calculator for anything. I'm being able to do this all through skills. So if I want 26, we're going to look at the six and 20, and I'm going to say X + 6 / 20 and X plus. Oh wait, just did that all wrong. I was jumping ahead of myself. X + 6 over the leading coefficient of eight X + 20 / 8 reduce, we get X + 3/4 and we get X plus. Let's see, a four goes into each of these, so 5 halves. Once I've reduced the fraction, if there's a denominator, I put it in front of the variable for its leading coefficient. Four X + 3 * 2 X +5. If I foil it out, I get eight X ^2 + 20 six X + 15. This next example, I needed the sum to be a negative number, but I needed it to multiply to give me a positive. So the way we do that is we know a negative times a negative gives us a positive. So when I do the sum here, these would all be negative numbers along with all these being negatives. So I went -23 so -8 and -15 So X -, 8 / 8 and X -, 15 / 8 we reduce it. X - 1 X -15 eighths won't reduce. So anything that still has a fraction comes out as the leading coefficient. One more example we want -37 if I look in this list, do I ever see a negative 37? No, I don't. Well, what does that mean? It means I can't write it as a binomial times binomial. The only thing I could do here is to factor out of one. So one times eight X ^2 -, 37 X -15. Now on the next page, I used 8 and 15 again to let you have some practice using the column we already made of our numbers. But you might want to see if you can come up with this column on your own just to get some more practice. And then at the bottom, we gave you 12 and seven once again, get some more practice with some different numbers. Thank you and have a wonderful day.