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CCSS.Math:

so we have an equation right here that says negative 2y times the expression y squared plus CY minus 3 is equal to dy to the third plus 12y squared plus FY and what I'd like you to do is pause the video and see if you can figure out what the variables C D and F are alright so let's work through it so at first this is going to look kind of intimidate we just have an equation you're trying to figure out what these three variables are how do you do it and once one reasonable approach would be let's just try to simplify what we have here on the left hand side and to simplify that we can just distribute the negative 2y onto this polynomial onto the y squared plus CY minus 3 and then we can set it equal to what we have here on the right and let's see if we can match up coefficients so let's distribute it so first we can think about what negative 2i times y squared is going to be negative 2y times y squared well that's going to be negative 2y to the third power because Y to the first times y squared is y to the third power all right now let's distribute now let's multiply the negative 2y times cy well that's going to be negative 2 times C so negative to C and then you're going to have Y times y so negative to C Y squared and then we would multiply the negative 2y times we got keep in mind this is a we're subtracting 3 here so we could view that as a negative 3 negative 2y times negative 3 negative 2 times negative 3 is positive 6 and we still have that Y over there so there you have it we have simplified the left-hand side of this equation and now let's see if let me just actually just write the right-hand side in different colors and then things might jump out at us so over here I wrote this the the third degree term the y to the third term I wrote that in blue so let me write the Y to the third term here in blue as well so d y to the third and then I wrote the y squared term in magenta so let me write the y squared term in magenta here so plus 12y squared and then last but not least I wrote the first degree term this y term in green so let me write the first degree term in green right over here so plus F why when you see it like that you see which terms match up which with which with which terms match up to which other terms so we could see that look I have the third degree term here that has to match up to this third degree term there so D D needs to be equal to negative two so let's write that down D is equal to negative two we can see the second degree term the second degree term matches up to this second degree term right over there so that tells us that this coefficient the 12 must be equal to the negative to C so let's write that down negative to C is equal to 12 to solve for C we can divide both sides by negative 2 and we get C is equal to 12 divided by negative 2 is negative 6 and that makes sense if C is negative 6 negative 2 times negative 6 or subtracting 2 times negative 6 is subtracting negative 12 which would be the same thing as adding 12 so then we got we get the exact same coefficient for the second degree term I think you see where this is going over here we have a 6y on the first degree term here we have an F Y on the first degree term this F must be the same thing must be equal to the coefficient here so f must be equal to 6 and we're done and the key realization here is you you match up the corresponding degree terms there's no way that well I don't want to get too too complex here but the simplest way to address this mystery is okay I have a third degree term here have a third degree term here well just if I look at it very simply the way I'm going to get this third degree term here is using this third degree term and I just look at the coefficients and say ok they must have the same coefficient and then we say D is equal to negative 2 and then we keep doing that