WEBVTT mathematics/algebra-1/smith
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Welcome back to www.educator.com.
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In this lesson we are going to take a look at factoring some trinomials using a method known as AC method.
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We have already seen factoring trinomials once before but these ones are going to be a little bit more complicated
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and that our squared term will be something other than the number 1.
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This is why we are going to pick up the AC method, we have a little bit more algebraic way to approach these types of problems.
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Recall some of the earlier trinomials that we have been factoring so far.
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The squared term in front has always been 1 and that made life pretty easy on us
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because when we went searching for those two binomials to break it down into.
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We did not have a lot of options in order to get that first term.
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It is probably something like y and y or x and x.
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There is not a whole lot of other things it could be.
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The reason why this made things a little bit easier is we only have to focus on the last term
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and making sure our outside and inside terms combined to give us the middle term.
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Now, we do not want to necessarily stick with those types of trinomials.
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We want to go ahead and factor things where the initial term is something other than 1.
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Now this will end up making things a little bit more difficult.
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The good news is with these ones you can still use something like the reverse foil method.
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You have not only possibilities for your first term, but now you also have possibilities for your first and last terms, both of those.
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This will make it a little bit more difficult when we are checking to make sure that the outside and inside terms combine to give us that middle term.
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Let us do a reverse foil example, so you can see that we have to track down many more possibilities.
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This one is 2x² + 7x + 6.
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It is not that big but we will look at the two binomials that we are looking to break this down into.
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Like before, I will be looking for two terms that multiplied to give us the 2x².
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Since that number is there I have to look at possible things that will give us 2.
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This one is not too bad, it has to be 1 and 2 to get that x², 1x and 2x.
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We will look at the possibilities that will give us our second term.
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6 could be 1 and 6, could be 2 and 3, or it could actually be those values flipped around.
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The question is what should it be? What things are we looking for here?
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I know that these two numbers whatever they are looks like they both better be positive since these are both positive.
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It could be 1 and 6, could be 2 and 3.
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Let us go ahead and put one of those in there just to see what happens.
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Let us suppose I'm trying out 1and I'm trying out 6.
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My first terms multiply out just fine.
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My last terms multiply out just fine.
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Let us check out our outside and inside terms.
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The outside would give us 6x and the inside would give us 2x.
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When you combine those together, you get 8x which unfortunately is not the same as that middle term.
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I need to come up with some other choice for those last terms.
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Let us see, if it is not 1 and 6, I guess we have to try 2 and 3.
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I will put those in there and we will double check our outside and inside terms.
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Outside would be 3x, inside would be 4x and sure enough those do combine to give us that 7x.
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The important part to recognize is that if your initial term is something other than 1,
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you have to look at your possibilities for your last term and your first term
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and play around with how they are ordered in order to get your proper factorization.
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Let us try another one that has a few more possibilities for that first term just to make things a little bit more interesting.
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Looking at the first term, we need something that will multiply and give us 6y².
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That could be 1 and 6, could be 2 and 3, but they both will definitely contain y because of that y².
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Looking at our last terms, many different things could multiply and give us the 10.
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1 and 10, 2 and 5 or possibly those just reversed around.
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I have lots of different options that I can end up packaging this together.
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Maybe this is 1 and 6 with the 1 and 10 or maybe I should use the 2 and 3 with the 1 and 10,
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or the 1 and 6 with the 2 and 5, or the 2 and 3 with the 2 and 5.
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There are many different ways and I can also do these in different orders to make sure that they combine.
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The key for figuring out which combination should you use is looking at those outside and inside terms.
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In this one, we want them to combine to give us that 19.
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We definitely have to do a little bit of work to figure out what that is.
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Let us see for the current setup my outside would be 10y and my inside would be 6y,
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which unfortunately does not combine enough to give us that 19y in the middle.
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I know that my 1 and 6, and my 1 and 10 I need to change something around this one.
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This one is just not going to work that way.
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Let us play around with our first term.
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Let us try something else for the beginning here.
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Let us try 2 and 3.
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What will that give us?
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I can see the outside is 20y, the inside is 3y but that is a little too much, 23y not going to work out.
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The good news is I did do this one earlier so I do have a combination that will actually factor.
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What we are looking for this one is 2, 3, 5 and 2.
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Let us check the outside and inside terms for this guy.
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The outside would be 4y and the inside would be 15y and sure enough those combine to give us 19y.
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I know that this is the proper factorization.
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Notice what this highlight is that when your initial term is something other than 1, and you have lots and lots of possibilities to run through,
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it can be very difficult to find out just the right combination of numbers to use in order to make it all work out correctly.
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If my numbers were even bigger I would have even more possibilities to run through.
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This is a problem.
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To fix this problem where our leading term is something other than 1, and our numbers could get fairly large,
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we do not necessarily want to use the reverse foil method.
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There are simply too many possibilities to consider for some problems and it gets too difficult.
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This is why I run to pick up something known as the AC method.
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This is a little bit more of an algebraic method that we can use and hunt down some of the possibilities we need for breaking it down.
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Let me quickly run you through how the AC method works, and then I will give you some quick tips on using it.
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The very first thing that you want to do when using AC method is just to see if everything has a common factor or not.
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And if it does have a common factor, go ahead and factor that out before beginning any other type of factoring process.
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If they have anything in common, pull that out.
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Then I will multiply the first and the last terms together, this is known as the A and C terms.
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This is where the method gets its name.
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Once you get that new number, you will be looking for two numbers that multiply to give you AC and they actually add to get you B.
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This will have the feel of looking for those two integers, but it will be a little bit more straightforward than what you see in the past.
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I do have a nice way to organize that step to keep track of the two numbers are looking for.
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Here comes the interesting part, when you find those two numbers we will actually split up your middle term into two new numbers
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and then you will have four terms total and you will actually use factor by grouping to move from there.
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The AC method is a way of splitting up your middle term and using a different approach that factor by grouping to handle it instead.
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Let me give you some tips on using the AC method.
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When you use the AC method you want to organize where your AC and your B terms go.
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What I recommend is that you draw a small x.
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In the top part of that x you put the values of A× C and in the bottom part of that actually put the value of B.
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What you are looking to do is you want to fill out the rest of this x by putting in these two numbers.
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The numbers that will go there, they must multiply to give you that top number and they must add to give you the bottom number.
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You will feel like you are filling out a very small crossword.
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Be very careful in doing this and make sure that the signs matchup.
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If you need to add to give you a negative number then make sure the two side numbers will add to give you that negative number.
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Always be careful on your signs with this one, they should matchup.
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You have heard a lot about the AC method and have not seen it yet, let us go ahead and do an example see can see it all in action.
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We want to factor 10q² – 23q +12.
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This is a good example of one that you want use the AC method on.
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If you try to factor directly you have lots of possibilities for the first term, like 1 and 10, 2 and 5.
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You have lots of possibilities for the second term, 1 and 12, 2 and 6, 3 and 4.
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Of course all of those reversed.
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Let us tackle this using the AC method.
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In step one, check all of your numbers to see if they have a common factor.
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I get 10, 23 and 12 it looks like they do not have anything in common.
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Unfortunately, means I can not pull anything out and make the number smaller.
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In to step two, I want to go ahead and multiply my a term and my c term together.
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10 × 12 = 120.
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I’m looking for two numbers that will multiply to give me 120 and add to be -23.
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This is where little box will come in handy.
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Let me just put in our few little notes.
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We want them to multiply that gives us our top number.
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We want them to add to give us that -23.
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To help us better find the numbers that will go ahead and do this,
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I’m going to start listing out all the pairs of numbers that will multiply to give us 120.
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1 and 120, 2 and 60, I have 3 and 40, it just keep continuing making this list until you get as many numbers as possible.
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I got 4 and 30, 5 and 24, 6 and 20, 8 and 15.
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Now that we have a bunch of numbers on this list, let us see how we can use it.
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We want these two numbers to multiply to be 120 and when we built those list that all should multiply to give us 120.
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But they must add to give us -23.
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The only way you want to add to get a negative number and multiply to get a positive number
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is if both of these new numbers here were negative.
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Let us say if you are going to pick two things off this list that will give us -23 when added together.
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I think it is going to be that last two, 8 and 15.
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Notice how those will multiply negative × negative will give us that positive 120 and will definitely add to be -23.
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Those are the two numbers we want.
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Now comes the interesting thing.
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What I'm going to do with those two numbers is end up splitting up my original middle term.
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I'm writing down the numbers of my original polynomial but I'm not writing that -23 in there.
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This is where I split it into two terms.
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This will be -8q – 15q so I have not changed my polynomial.
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I just take a look at it in a different way.
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You will notice how this new polynomial has four terms 1, 2, 3, 4.
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I'm going to now attack it using factor by grouping, which means I will take these terms two at a time.
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Let us do the first two and see what they have in common.
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Both are divisible by 2q.
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Let us see what we got left over in here.
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2q, I have 5q - 4 and let us see what is in common with the next two.
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It looks like I can pull out -3 from both of them.
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That would leave us 5q - 4 and notice how the signs do match up, -3 × 5 =-15.
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-3 × -4 =12.
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We can go ahead and wrapped this one up.
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They both have a 5q - 4 in common, I will write that for my first binomial with the leftover pieces being 2q and – 3.
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It is quite a journey to get to those final two binomials, but notice how it is a little bit more methodically,
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not necessarily guessing or picking things out of here.
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You have a better hunting way of going about it.
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Let us quickly check to make sure that this is the correct factor polynomial just by running through the foil process.
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5q × 2q = 10q², outside terms -15q, inside terms – 8q, last terms +12.
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It is already starting to look pretty good since it looks like that one.
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Just combine my little terms here and I get 10q² – 23q + 12, which is exactly the same as I had originally.
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I know that this one is factored correctly using that AC method.
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Let us see the AC method again just we can get more familiar with it.
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In this one we want to factor 5t² + 13t – 6.
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I want to make sure that they have a greatest common factor of only one,
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which means do they have anything in common that I can factor out at the very beginning.
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5, 13, and 6 do not have anything in common.
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Let us move on to multiplying the A and C terms together.
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5 × -6 = 30.
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Two numbers that will multiply to give us 30, but add to give us a 13.
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We will use our box to help us out.
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It need to multiply to be 30 and add to be 13.
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The two numbers that we put in here, they must multiply to give us 30 and they will add to give us 13.
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To help out with the search, we will list down all the things that multiply to give us 30.
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1 and 30, 2 and 15, 3 and 10, 4 and 15.
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The only two things that are going to work from this list or the least that I can see I think will be our 3 and 10.
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Let us go ahead and put those in.
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Hold on, I think we forgot one of our signs here should be -30 on top.
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We will try this again.
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We need two numbers from our list that will multiply to give us -30, but add to be 13.
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I think the 2 and the 15 will have to be the one to do it because they have to be different in sign to multiply to give us that -30.
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How about 15 and -2?
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We are ready to split up our middle term.
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I have written the first term and the last term now we will write out that middle term split using these two new numbers.
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– 2t + 15t looking pretty good.
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Now that we have this, we want to factor by grouping.
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We will take these two at a time.
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5t – 2t they only have one thing in common and that would be t.
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Let us see what is left over.
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5t - 2 looking at the next two terms 15t – 6 they have a 3 in common.
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5t – 2.
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We have the 5t -2 common piece, we will go ahead and take it out of both of them and write out our leftover pieces, t + 3.
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This one did take quite a bit of work let us quickly check it again by our formula.
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5t² + 15t – 2t -6.
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These two little terms combine giving us 13t - 6, which is the same as the original.
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I know that this is the correct factorization.
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Let us go ahead and try to factor this one using the AC method.
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This one has a few more variables in it so notice I have x² and y².
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Do not worry too much about those x and y, what you will see just focus mainly on those numbers and hunt down what those need to be in.
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Is there anything I can take out at the very beginning?
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Do they have anything in common?
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It looks like the 11 is going to make it where they do not have anything in common.
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I will multiply the A and C terms together.
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6 × -10 = -60.
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We need two numbers that will multiply to be -60, but add to be 11.
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Let us try our box to help us out.
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-60 and 11.
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They need to multiply and give us -60 and add to be the 11.
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Starting off with writing down all the possibilities to give is that 60.
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1 and 60 would do it, 2 and 30, 3 and 20, 4 and 15.
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The two numbers that we use must multiply to be -60.
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We want one of these numbers to be positive and the other one to be negative.
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Since we are adding to be 11, the larger number must be positive.
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I think I see a good option on this list, the 4 and the 15.
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The 15 is the larger one, so it must be positive and the 4 is the smaller one, so we will make it negative.
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We will write down our polynomial and split up the middle term into two new terms.
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We will use 15xy and - 4xy.
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Notice that we are using that xy here because those are what is on the middle term.
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Onto that factor by grouping process.
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We will look at our first two terms and we will look at our second two terms.
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Starting with the first two, what do they have in common?
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They both have a 3x but I can go ahead and draw out.
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That would leave me a 2x + 5y.
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Looking at the next two terms, they have a -2y in common.
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What would that leave?
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2x + 5y.
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Double check and make sure your signs will get -2 × 2 =-4, -2 × 5 =-10.
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Things are looking good.
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Okay, now I have that common piece we will take it out from both of them.
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2x + 5y and then 3x -2y.
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It looks like this one is factored and you can double check by running through the foil process.
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After all we get a little bit more familiar with that AC method in seeing the steps as you go along.
00:24:52.500 --> 00:25:00.400
Keep your components, find those two new numbers so you can split up that middle term.
00:25:00.400 --> 00:25:07.900
Let us do one last example and I want to do one where they had a greatest common factor that you could pull out.
00:25:07.900 --> 00:25:13.700
You will notice how initially these numbers are quite large, we have 28, 58 and -30.
00:25:13.700 --> 00:25:16.600
Furthermore, they all have an x² in common.
00:25:16.600 --> 00:25:23.400
Before we even start some of the AC method, let us go ahead and pull out that common term first.
00:25:23.400 --> 00:25:25.800
What do they have in common?
00:25:25.800 --> 00:25:29.000
Everything is divisible by 2 and they all have an x².
00:25:29.000 --> 00:25:34.300
Let us take out a 2x².
00:25:34.300 --> 00:25:37.600
Let us see if this makes things a little bit smaller.
00:25:37.600 --> 00:25:56.900
14x² – 58 ÷ 2 = 29x and this one divided by 2 – 15.
00:25:56.900 --> 00:25:59.000
It looks like they do not have anything else in common.
00:25:59.000 --> 00:26:02.200
We need to continue factoring it.
00:26:02.200 --> 00:26:09.300
From here I’m going to take its A and C term and I’m going to multiply those together.
00:26:09.300 --> 00:26:12.700
Now these ones are quite large, but we can do it.
00:26:12.700 --> 00:26:21.600
14 × -15 = -210 quite large.
00:26:21.600 --> 00:26:29.100
I need two numbers that multiply to give me -210, but they add to be -29.
00:26:29.100 --> 00:26:34.600
Let us draw our box and start hunting down some possibilities.
00:26:34.600 --> 00:26:41.800
I have -210, must add to be -29.
00:26:41.800 --> 00:26:45.100
I want to make this easy on ourselves, reduce it as easy as possible.
00:26:45.100 --> 00:26:49.600
I’m writing down possibilities that will multiply to be 210.
00:26:49.600 --> 00:27:13.000
1 and 210, 2 and 105, 3 and 70, 5 and 42, 6 and 35, 7 and 30, 10 and 21, 14 and 50.
00:27:13.000 --> 00:27:20.500
Now that I have a list of bunch of different numbers, we need to multiply to give us -210.
00:27:20.500 --> 00:27:26.900
That means one of these numbers will be negative and one of them will be positive.
00:27:26.900 --> 00:27:32.800
They will add to be -29 so I know the larger number must be a negative.
00:27:32.800 --> 00:27:37.200
It is the only way we will get -29 when adding.
00:27:37.200 --> 00:27:52.200
Come over the list very carefully, the one that will do it will be this pair right here, the 6 and 35.
00:27:52.200 --> 00:27:57.700
Sure enough, those multiply to give us the -210 and they add to give us -.29.
00:27:57.700 --> 00:28:05.600
Sometimes you might have to go through and check in these one by one, but it is worthy process.
00:28:05.600 --> 00:28:12.900
We have that and we are going to write down the 14x² - 15.
00:28:12.900 --> 00:28:23.400
Let us take our middle term and split it up 6x - 35x.
00:28:23.400 --> 00:28:27.700
Right now, we can continue with our factor by grouping.
00:28:27.700 --> 00:28:32.400
Grabbing this first two and looking at what they have in common.
00:28:32.400 --> 00:28:39.000
I see that we can take out 2x.
00:28:39.000 --> 00:28:49.500
That would leave us with 7x + 3.
00:28:49.500 --> 00:28:51.800
Looking at the next two numbers.
00:28:51.800 --> 00:29:02.300
These ones we can pull out -5.
00:29:02.300 --> 00:29:10.400
That would leave us with 7x and 3 looking pretty good.
00:29:10.400 --> 00:29:15.700
Now I can grab my two binomials and almost be done.
00:29:15.700 --> 00:29:19.600
They have a 7x + 3 in common.
00:29:19.600 --> 00:29:23.500
There would be 2x - 5 left over.
00:29:23.500 --> 00:29:26.200
Be careful this one is not done.
00:29:26.200 --> 00:29:34.100
Remember that initial factor we took out at the very beginning, it is still out front of this entire process.
00:29:34.100 --> 00:29:38.000
Feel free to write it down now in front of all of this.
00:29:38.000 --> 00:29:41.000
That way you do not forget it.
00:29:41.000 --> 00:29:48.300
Now we have the final factored form of our polynomial.
00:29:48.300 --> 00:29:54.500
Using the guess and check method and the AC method can be two great ways to factor your polynomials.
00:29:54.500 --> 00:30:01.100
I suggest using the reverse foil method if the numbers are not that big or the coefficient is 1.
00:30:01.100 --> 00:30:07.100
It starts to get a little bit more complicated then feel free use this AC method to break it down.
00:30:07.100 --> 00:30:09.000
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