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# Multiplying binomials intro

CCSS.Math:

## Video transcript

let's see if we can figure out the product of X minus 4 and X plus 7 and we want to write that product in standard quadratic form which is just a fancy way of saying a form where you have some coefficient on the second degree term ax squared plus some coefficient B on the first degree term plus the constant term so this right over here would be standard quadratic form so that's the form that we want to express this product in and I encourage you to pause the video and try to work through it on your own alright now let's work through this and the key when we're multiplying two binomials like this or actually when you're multiplying any polynomials it's just to remember the distributive property that we all by this point know quite well so what we could view this as is we can distribute this X minus 4 this entire expression over the X and the 7 so we could we could say that this is the same thing as X minus 4 times X plus X minus 4 times 7 so let's write that so X minus 4 times X or we could write this as x times X minus 4 that's distributing the or multiplying the X minus 4 times X that's right there plus plus 7 times X minus 4 times X minus 4 notice all we did is distribute the X minus 4 we took this whole thing and we multiplied it by each term over here we multiplied X by X minus 4 and we multiply 7 by X minus 4 now we see that we have these I guess you'd call them two separate terms and to simplify each of them or to multiply them out we just have to distribute in this first we have to distribute this blue X and over here we have to distribute this blue 7 so let's do that so here we could say x times X is going to be x squared x times we have a negative here so we could say negative 4 is going to be negative 4x and just like that we get x squared minus 4x and then over here we have 7 times X so that's going to be plus 7x and then we have 7 times the negative 4 which is negative 28 and we are almost done we can simplify it a little bit more we have two first degree terms here if I have if I have negative four X's and to that I add seven X's what is that going to be well those two terms together and these two terms together are going to be negative four plus seven X's negative four plus plus seven negative four plus seven X's so all I'm doing here are making it very clear that I'm adding these two coefficients and then we have all the other terms we have the x squared x squared plus this and then we have and then we have the minus and then we have the minus 28 and we're at the home stretch this would simplify to x squared now negative four plus seven is three so this is going to be plus three X that's what these two middle terms simplify to to 3x and then we have minus 28 - 28 and just like that we are done and a fun thing to think about a notice it's in the same form if we were to compare a is 1 B is 3 and C is negative 28 but it's interesting here to look at the pattern when we multiply these two binomials especially these two binomials where the coefficient on the x-term was a 1 notice we have x times X that's what actually forms the x squared term over here we have negative 4 let me do this in a new color we have negative 4 times that's not a new color we have we have negative 4 times 7 which is going to be negative 28 and then how did we get this middle term how did we get this 3x well you had you had the negative 4x plus the 7x or you had the negative 4 plus the 7 times X you had the negative 4 plus the 7 plus the 7 times X so hope you see a little bit of a pattern here for multiplying two binomials where the coefficients on the X term are both 1 it's going to be x squared and then the last term the constant term is going to be the product of these two constants negative 4 & 7 and then the first degree term right over here it's coefficient is going to be the sum of these two constants negative four and seven now this might you could do this pattern if you practice it it's just something that'll help you multiply binomials a little bit faster but it's super important that you realize where this came from this came from nothing more than applying the distributive property twice