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CCSS.Math: , ,

we're going to learn to recognize and factor perfect square polynomials in this video so for example to have the polynomial x squared plus 6x plus 9 and if someone asks you hey can you factor this into two binomials well using techniques we learn in other videos okay I need to find two numbers whose product is 9 and whose sum is 6 so I encourage you to think of to pause this video and say well what two numbers can I can add up to 6 and if I take the product I get 9 well 9 only has so many factors really 1 3 & 9 & 1 plus 9 does not equal 6 and so and negative 1 plus negative 9 does not equal 6 but 3 times 3 equals 9 and 3 plus 3 does equal 6 3 times 3 3 plus 3 and so we can factor this as X plus 3 times X plus 3 which is of course the same thing as X plus 3 squared and so what was it about this expression that made us recognize or maybe now we will start to recognize as being a perfect square well I have of course some variable that is being squared which we need I have some perfect square as the constant and that whatever is being squared there I have 2 times that as the coefficient on this first degree term here let's see if that is generally true and I'll switch up the variables just to show that we can so let's say that I have a squared plus 14 a plus 49 so a few interesting things are happening here all right I have my variable squared I have a perfect square constant term that is 7 squared right over here and my coefficient on my first degree term here that is 2 times the thing that's being squared that is 2 times 7 or you could say it's 7 plus 7 so you can immediately say okay if I want to factor this this is going to be a plus 7 squared and you can of course verify that by multiplying out by figuring out what 8 plus 7 squared is sometimes when you're first learning is okay isn't that just a squared plus 7 squared no remember this is the same thing as a plus 7 times a plus 7 and you can calculate this by using the foil fo IL technique I don't like that so much because you're not thinking mathematically about what's happening really yourself to do this tribute of property twice here first you can multiply a plus seven times a so a plus seven times a and then multiply a plus seven times seven so plus a plus seven times seven and so this is going to be a squared plus seven a plus now we distribute this seven plus seven a plus forty-nine so now you see where that 14 a came from its from the seven a plus the seven a you see where the a squared came from and you see where the 49 came from and you can say speak of this in more general terms if I wanted to if I wanted to just take the expression A plus B and square it that's just a plus B times a plus B and we do exactly what we did just here but here I'm just doing in very general terms with a or B and you can think of a is either a constant number or even a variable and so this is going to be if we distribute this it's going to be a plus B times that a plus a plus B times that B and so there's going to be a squared now I'm just doing the distributive property again a squared plus a B plus a B plus B squared so it's a squared plus two a B plus B B squared so this is going to be the general form so if a is the variable which is X or or a in this case then it's just going to be whatever squared in the constant term is going to be two times that times the variable and I want to show that there's some variation that you can your tain here so if you were to see 25 plus 10 X plus x squared and someone wanted you said hey why don't you factor that or you can say look this right here is a perfect square it's 5 squared I have the variable squared right over here and then this coefficient on our first degree term is 2 times 5 and so you might immediately recognize this as 5 plus x squared now of course you could just rewrite this polynomial as x squared plus 10x plus 25 in which case you might say okay variable squared some number squared 5 squared 2 times that number is the coefficient here so that's going to be X plus 5 squared and that's good because these two things are absolutely equivalent