all right so we're gonna need to be able to factor for the next lecture so we're looking at the first example of a factoring technique which is difference of two squares so if I'm subtracting one square from another it can be factored into this form here A minus B times a plus b so it's important that we make sure that we figure out what are the things that we are squaring in each of our terms um so for the first example we have x squared minus 25 all right and I want to figure out how I could rewrite this so it looks more like a squared minus B squared well the x squared looks good but I gotta remember that 25 is the same thing as 5 squared all right and then this x squared minus 5 squared now looks like this a squared minus B squared all right and so we can Factor it into this form here and we end up with x minus 5 times x Plus 5. all right so in our second example it looks a little different because we've got this coefficient of 9 on there not a big deal we know that 16 is the same thing as squaring 4. and so now what I need is a quantity that's being squared because I need a squared minus B squared so the number that I Square to get 9 is 3 but I'm also squaring the Y so that's what's in the parentheses there so this really means 3 squared times y squared which is 9y squared and so now my a in this case is 4 and my b is 3y and so this factors into 4 minus 3y times 4 plus 3y all right so as long as you know the formula for the difference of two squares shouldn't be too difficult right let's go on to factoring trinomials so trinomials when we Factor these at the coefficient on x squared is one they're typically pretty straightforward what I need are two numbers that multiply together to give me seven but add together to give me eight or sorry excuse me negative eight and the two numbers that would do that would be negative seven and negative one because when you multiply two negatives you get a positive and negative seven minus one is negative eight so when this factors it becomes x minus seven and then x minus one and again you can always check these by Distributing everything through um many of you guys remember the acronym foil that only works when you have a first and outside and inside and a last so you'll not hear us use that term very often now when the coefficient on x squared is not one it's a little bit more difficult there's a little bit more work to be done so we're going to use What's called the AC method which is similar to what you see up here remember this is going to be a x squared plus b x plus C the AC method is literally what it says we're going to multiply a which is 2 times C which is 5. so we're gonna do that over here and we know that 2 times 5 is 10. all right and just like in the above problem we need two numbers that multiply together to give me 10. but add together to give us 11. okay and that's pretty straightforward that's just going to be ten and one but that doesn't mean I can just factor and jump in like I did earlier because with the coefficient of 2 on there it doesn't work that way what we're going to do is take 11 x and rewrite it as 10x plus 1X it's still 11x it's just broken down and then from this point we are going to go ahead and Factor by grouping so I'm going to look at the first two terms together in a group Plus the second two terms together in a group and in the first group I notice that I have a greatest common factor of 2 times x so I'm going to go ahead and factor out a 2X that'll leave me with a 1x in the first term and then 2 coming out of 10 leaves me with 5. there's nothing I can factor out of the second group of terms but that's okay because I see that I have a common binomial I have in my first term I have a binomial of X plus 5 and in my second term I have a binomial of X plus five so I can now factor that binomial out and that will leave me with 2x in my first term but I'll only be left with one in my second term and so now I've finished factoring with the AC method all right next thing we're going to need to be able to do is understand conjugates and how to multiply them earlier though we looked at the difference of two squares notice this a minus b and a plus b are conjugates so what basically we're looking at is the opposite operation okay when we look at the product of something with its conjugate it just gives us difference of two squares it's a beautiful relationship all right so we're going to identify the conjugates for these two expressions and then we're going to multiply by that conjugate okay so I have x minus six so its conjugate would be X Plus 6. all right and so when I multiply this out I can again use the fact that x minus 6 times x plus six will just be x squared minus 6 squared because again when you multiply something by its conjugate you just get difference of two squares and then we can obviously clean that up a teeny bit by squaring the six and we have x squared minus 36. all right now I have 4 plus the square root of Y over here so its conjugate would be 4 minus the square root of Y just the opposite operation all right and just like before we have our difference of two squares factored form so when we multiply this together we're going to get the first term squared so 4 squared minus the second term squared so the square root of Y squared and we know 4 squared to be 16 and we also know that when we square a square root they cancel out and we're left with just in this case why so when you multiply um an expression by its conjugate you're just performing the difference of two squares multiplication and we will need this quite a bit in the next section of the textbook the next thing we're going to look at is complex fractions there are several ways that we can go ahead and simplify these things we don't like the the way they look at the moment because we have fractions within fractions rational expressions within rational expressions and they can be simplified quite a bit so what we're going to do is use a technique that makes this a little bit easier to get through a little bit quicker to get through we're going to look for all of the denominators that we're trying to I don't want to say eliminate but we're trying to we're trying to get these out of fraction form so I see that I have x minus 1 in both of these terms as a denominator and I have X in this term as a denominator so what I want to do is multiply this entire fraction all right this entire rational expression by the X here because I want to multiply 1 over X by X so it cancels out and I want to multiply by x minus 1. but whatever I multiply into the numerator I must also multiply into the denominator so I have to be careful to do that both in the numerator and the denominator from this point you're going to need to distribute this to all terms and so now what we have and I'll write an I'll write an intermediate step here that maybe you won't need in the future I have 2 over x minus 1 but I'm multiplying that numerator Now by x times x minus 1. and in my next term I have 1 over X but now I'm multiplying it by x times x minus 1. and then in my denominator I have 3 over x minus one but again I'm multiplying it now by x times x minus 1. and again this may be a step you don't need to show if you can do this the next thing in your head which is to cancel you know common factors from numerators and denominators see this I have x minus 1 and I have x minus 1 that cancels here I have x divided by X that's going to cancel and then lastly I have x minus 1 divided by x minus 1 which is going to cancel all right now when I finish this and I start cleaning it up I have the 2x plus I'm left with x minus 1 here and then in the denominator I'm left with 3x all right and then I can combine some like terms here and I have 3x minus 1 all over 3 x and then I'm just going to make a quick statement I'm going to return back to the beginning and say well at the very beginning of this problem if x was equal to 1 these two terms would have been undefined and then therefore this entire complex fraction would have been undefined so I know that X cannot be one I also know that X cannot be zero all right that's already shown here but it doesn't hurt for us to just make sure that we make the statement that X cannot be one and X cannot be zero because if it is equal to either of those two values the entire expression is undefined so we'll do one more example of this and over here I see that I actually have three different denominators and I want to be careful not to go overboard on this all right I know I need to multiply by X Plus 2. that's absolutely necessary that's a little messy sorry got a little snug over there guys um but then I know I need to multiply by X to get rid of this denominator but if I multiply by x squared to get rid of this denominator it'll take care of that so I don't need to multiply by both x and x squared okay that's not going to be necessary the least common denominator is all we need to multiply by same thing we're going to go ahead and distribute everything through all right and so now we have in our first term x times x squared times X plus 2 all over X plus 2 again this is that step that you might be able to skip over 4 times x squared times X plus 2 all over X all right and then in my denominator we have one times x squared times X Plus 2. all divided by x squared and then again our goal here is to go ahead and cancel whatever we can so in the first term I can cancel X Plus 2. in the second term x squared divided by X will bring that down to x to the first in the numerator and then down here x squared divided by x squared those will cancel out as well so now we have x cubed is what's left here because of x times x squared minus we have 4X times X plus two all divided by X plus 2 is all that's left down there in the denominator so the last thing I'm going to do is distribute the negative 4X into the binomial X Plus 2. that'll give me negative 4x squared minus 8 X 4 times 2 is 8. all of that will be over X Plus 2. all right and then the last thing again make sure that we go through and do this the two values for X that will make our original expression undefined I think zero is a pretty clear one here and then the value of x equals negative two would be a problem as well right if you feel like you need more um review on these three topics there are three separate videos they are all longer videos they can be found in my courses um probably a little bit later on in the week but that should help you guys with getting ready for the next lecture