a substitution so i will i'm just going to say that t equal that NUMBER_TOKEN to the x so this equation becomes t to the third power plus t to the second power plus t and then minus NUMBER_TOKEN and that's equal to zero okay how do we solve this cubic equation um we cannot factor this not even by grouping right so we will just have to hope that it does have some nice answers meaning rational answers and we will have to use the rational NUMBER_TOKEN. meaning that let's just go ahead and look at the factors of negative NUMBER_TOKEN which are plus or minus NUMBER_TOKEN plus or minus NUMBER_TOKEN plus minus NUMBER_TOKEN and also plus or minus NUMBER_TOKEN. and of course this is just one so it's pretty nice and we're just going to plug in all these numbers into the t's to see if it works out or not and of course you can just verify it on your own will see that when t is equal to NUMBER_TOKEN it works so just say notice that t is equal to NUMBER_TOKEN is a solution after you verify it on your own so once we know this we know this right here can be factored into t minus NUMBER_TOKEN times something else and this right here is just going to be a quadratic which is so nice how do we figure this out well we can do polynomial division or synthetic diffusion we will do the second one of course so here we go let me draw the box like this and put on the coefficients right here we have NUMBER_TOKEN NUMBER_TOKEN NUMBER_TOKEN and then minus NUMBER_TOKEN. and then we're just going to put down two right here and here we go bring down the one right here and we do two times one which is two we put it here and then we add this and that which is three and then two times three is six and put it down right here NUMBER_TOKEN plus NUMBER_TOKEN is NUMBER_TOKEN and then NUMBER_TOKEN times NUMBER_TOKEN is NUMBER_TOKEN we put it here and you see this plus that just NUMBER_TOKEN and of course we should be getting NUMBER_TOKEN for the remainder because we know that this right here divides that cubic polynomial and now for this right here we have the coefficients here already originally it was t to the third power so this right here is the coefficient for t squared this is going to be the coefficient for t to the first power and this is just a constant term so this right here is just going to be NUMBER_TOKENt squared plus NUMBER_TOKENt and then plus NUMBER_TOKEN. so now we can just take up this and of course we know t is equal to NUMBER_TOKEN already so let's just really focus on that from here we can say t squared plus NUMBER_TOKENt plus NUMBER_TOKEN is equal to NUMBER_TOKEN we can just use the quadratic formula so we get t is equal to negative b which is NUMBER_TOKEN and then plus or minus square root of b squared which is NUMBER_TOKEN squared and then minus NUMBER_TOKEN a c a is NUMBER_TOKEN c is NUMBER_TOKEN and then all divided by NUMBER_TOKEN times a which is NUMBER_TOKEN times NUMBER_TOKEN. okay this is negative three and then plus or minus square root three minus NUMBER_TOKEN which is negative NUMBER_TOKEN and then divided by two and now we notice that we have a negative instead of a square root so we can just put that on outside make that into i so this becomes negative NUMBER_TOKEN plus or minus i square root of NUMBER_TOKEN all over NUMBER_TOKEN like this great but these are t's we want x's right so from here we know t is NUMBER_TOKEN that means NUMBER_TOKEN to the x is equal to NUMBER_TOKEN.
NO_SEGMENT_TOKEN