Algebra 2 Review for 6.1 – 6.4 Quiz

okay this is review for Chapter six and I think it’s six point one through six point four and we already covered one through five in class to prepare you for the quiz on Tuesday and so I will start with my other examples beginning with problem number six you did not see but the directions were just factored x cubed plus 7x squared plus 10x and maybe fourth hours so this one I’m not sure maybe whenever you’re asked to factor the factor you got a few choices but the first thing I’d always would do is check for a GCF so you look at each term it’s in standard form which is a great form to start with so highest degree is first and then we go and decrease in order execute x squared the next to the first and right away I’d notice I have an X that everybody has in common so I divide that X out I factored away and so each term has decreased by one degree then I look at what I have left and I’d say hold it that’s a quadratic trinomial and we always tried to break those up into two binomials remember this ten comes from the last term in two and five multiplied together those things multiply and give you a ten and then the other spot we look at is we look at this product combined with this product the outside and inside and 2x and 5x add up to 7x which is what I want in the middle actually want this 7x the x squared is taken care of by this X out here and so those are both positive and so our final factorization is the X and then the X plus 2 times X plus 5 and you could always check these things remember if you foil those together you get x squared plus 7x plus 10 if I distribute the X to everybody I would end up right back where I started so this and deed is the complete factorization I know I’m done because now I’ve got X to the first X to the first X to the first everybody is linear and you can’t factor it past that problem number seven looks like this it says divide x to the fourth – it’s a four up there for X cubed minus X minus four and you’re gonna divide that by X cubed minus one my preference is to use synthetic division whenever I can however because I’m dividing by a cubic I cannot use synthetic so I have to set up a long division problem and hope I have enough room on this screen to do it oh I’m getting ready to write down my next term I notice it’s X to the fourth X cubed there is no x squared so I’ve got to put in a filler for that as zero x squared is still 0 then it comes my minus X then my minus 4 so I’ve made sure what I’m dividing into my dividend is just going in decreasing order in standard form and then here’s my divisor execute -1 that’s missing a term as well X cubed + no X Squared’s plus no X’s minus 1 so I had to put in two zeros for that divisor so here we go we ask yourself what do I have to take x cubed times X to the fourth and the answer is an X so now I distribute that X to everybody to figure out what I’m gonna subtract and I golike x times X cubed I’m gonna get plus a 0 X cubes + 0 X Squared’s minus 1 times X is X this is the tricky part for most people now you have to subtract that entire thing so I’m doing my first my X to the fourth and I go X to the fourth minus X to the fourth is 0 now negative x negative 4 X cubed minus 0 is just negative 4x cubed 0 minus 0 is squared and negative X minus negative x is also 0 because it’s gonna be negative X plus X I bring down the negative 4 so I’ve got four terms because that’s what I’m dividing into it to get negative 4 X cubed I have to take this X cubed times a negative 4 and I get negative 4 X cubed I get – oops doesn’t matter if it’s minus or plus I’m taking it times a 0 and times another 0 guess I should put

0 X’s there so you can see what’s going on and negative 1 times negative 4 is plus full-line subtract negative 4 minus negative 4 is gonna be a plus so they cancel out 0 my 0 0 0 – 0 0 and I’m finally down to the just these last two negative 4 minus negative 4 is negative 8 there’s nothing left appear to bring down I’ve worked through all my terms so that’s my remainder so my answer is this is equal to X minus 4 with a remainder of negative 8 or more specifically now I’d be happy with that because we’re just gonna write things with remainders but it’s X minus 4 plus remainder over divisor X cubed minus 1 so that’s a fractional part that it goes into it okay so there was a long division problem remembering to fill in those zeros to keep all of my terms lined up problem number 8 another division problem X cubed minus 4x squared a 6x minus 4 divided by X minus 2 this is just a binomial with a linear our linear binomial rather coefficient of x is just a 1 out in front of it those are set up perfectly for synthetic division then remember you divide by a user positive 2 in the box and we’re always ask yourself what makes that zero and positive 2 does our coefficients let’s mark all those there’s one negative for no yep six and then negative for okay so it’s all the coefficients of so I’m one negative four six and then my constant to it draw a line step one bring down the 1 and then multiply and put the result up here two times one is two and then we add and then we multiply that 2 times negative 2 we get negative 4 and we add and 2 times 2 is 4 and we add now remember our final answer is not 1 negative 2 to 0 final answer is what’s the answer to this division problem I took a cubic divided about next divided out to next to the first so I’ve gone down by one degree I’m down to x squared minus 2x plus 2 and so there’s the answer to that division problem well quickly do that one off to the side here to show you long division and show you that does give exactly the same answer I’ve got X cubed minus 4x squared plus 6 X and minus 4 and so I’d go x squared and I distribute it and get X cubed minus 2 x squared subtract that get negative 4 plus 2 which is negative 2x squared and I bring down the 6 X next goes to negative 2 X gives me native 2x squared when I distribute plus 4 X subtract that 6 6 minus 4 is 2 X and I bring down the negative 4 and I multiply by a 2 to get 2x minus 4 I subtract my remainder is 0 and you can see the division gives me x squared minus 2x plus 2 don’t have to check these things obviously between the point of synthetic division just showing you that that’s what our answer comes out to be in long division as well question number 9 is x- for a factor of three x cubed plus 10x squared minus X minus 12 so basically that’s saying hey does X minus 4 divide evenly into that polynomial and I would check that by using synthetic with a four running down my coefficients remember there’s a 1 in front of that negative x so that negative 1 is right there I draw my line bring down my 3 multiply and add multiply and add and I really can’t believe that’s the correct answer 310 native 1 native 12 using 4 that’s 3 12 22 88 87 some really big number and some big number down here minus 12 my remainder is not 0 I don’t even have to know what it is because of the question is that a factor the answer

is new it’s not a factor it was a factor my remainder would have to come out to be 0 number 10 different type of question 10 is insane is something a factor it is saying factor this saying tell me how this thing breaks up is a product of stuff well it’s got two terms and the first one is a cube term so it’s probably gonna be a difference of two cubes so that’s a 3x that’s cubed – 3 cubed is 27 X cubed is X cubed + 8 is a 2 so it is a difference of two things that have been cubed you have to remember the formula for a difference of cubes a minus B times a squared plus a B plus B squared and so now it’s just a plug and chug problem we go a minus B a is the 3x in here and be is the – so it’s 3x – to take that a term the 3x and square it 3 squared is 9 and x squared squared is our x squared is x squared then a times B 3x x – 6 X and finally take the to the B value and squirt there’s the exact the factorization for it and this remember is always prime so you stop right here you have to write prime we just know it is 11 is kind of a continuation of that problem but it was gonna be easiest in class to show you both without having to really solve or do them both so 11 it says solve this now if you look back at number 10 10 is 27 X cubed – 8 and now we’re trying to solve it so now it says hey the equals 0 try and figure out what X has to be this one since we know what factors and we just saw the factorization for it I’m gonna write it down again since we have it factored all we have to do is set each factor equal to zero and solve each of those the one left this guy he’s easy because he’s linear with love linear factors there’s one of our solutions if you were to graph this you would see it has a zero where it crosses the x-axis at 2/3 that’s this solution that we just found however this other term our other factor here is prime it’s quadratic we can’t unfold it and the only way we can solve it is to use the quad formula on it remember this is a B and C so we have negative B B squared minus 4 times a times C all over two times a now the stuff underneath the radical that’s your radicand I’d be careful with that thing make sure you get that right C on your calculator 36 minus 4 times 9 times 9 negative 288 I think today in class I came out with a negative 108 using my head drop something there for AC and there’s my problem hold on a second C is incorrect you look back here at the factorization this last term oops this last term should’ve have 4 and I’ve put a 9 in there I’m not sure where my head was on that that should be yeah yeah 2 squared 4 it’s gonna change a few things for AC C is now a 4 in here and so 36 minus 4 times 9 times 4 yeah negative 108 there we go over 18 okay so this I’m going to copy this to the next page here take that I’ll copy it cuz my next job is after I figured out what that discriminant is is to simplify it okay so let’s walk through that I’m just gonna work with this square root of negative 108 then we’ll stick it back in the problem so I’ve pulled it out to the side nice take out the eye first okay so that’s just square root of 108 now I’m gonna take that 108 and I’m gonna start to factor it okay that’s two times 54 and I’m looking for perfect squares and I don’t see one yet so that’s 2 times 2

times 27 okay there’s that too and then 54 factored some more I can see now there’s a 2 by 2 in there so I could take out a 2 because the square root of 4 is 2 so I’ll just continue to factor it so I’m going to get it down to all of its prime 8:27 is 3s multiplied together three of them so now I’m ready to start taking stuff out of my radical oh that’s a four I can take the square root of that let’s do that’s a nine nine takes square root of that that’s a three so a two comes out the three comes out that’s six I have my eye outside and my three left in so now working back up here in the original solution that I’m trying to simplify its negative six plus or minus I’ve simplified the radical part to 6i root three over 18 that numerator has a GCF at six and I did that because I notice 18 has a six in it as well so I can do some canceling and I get negative 6 plus or minus I root 3 over 3 keep in mind that’s two solutions down here that’s negative 6 plus the I root 3 over 3 and negative 6 minus the I root 3 over 3 if you remember this was a cubic equation I should get three answers my two of them are right here remember that counts as two answers and then I did have one at the very beginning on this slide X is 2/3 so that’s my third solution okay so that’s completely solved now I’ve got three answers that I can put down question number 12 kind of a phrase little differently than we may be used to what was f of X 12 up here equals 4x cubed and G of x equals 4x squared plus 3x and the question was find X such that or so that however you want to phrase it or the X that makes it true find X so that f of X equals G of X so essentially what you’re trying to do is solve this equation okay so that’s the equation we’re trying to solve and this is where in your calculators our book gives us one choice and I suggested another one the book suggestion is suggestion is to graph this as y1 graph this is y2 and look where they intersect each other and my method was hey get it equal to 0 and then look for the zeroes of it where does it cross the x-axis ok for that you just say find zero on your calculator see if I have time I’ll try and shoot a video for that but right away one of the things I’m going to notice is hey if you want to solve this I know if there’s a GCF of X and then I would try to factor this and I’m gonna try to X and 2x I’m going to try a 3 and a 1 there’s 6 this is 2 I want to get a negative 4x in the mail so I want my 6x negative and this one positive okay so if I were to graph this thing on my calculator it should show me finding each of my zeros x equals 0 to X minus 3 equals 0 so I’m setting each factor equal to 0 that’s each of these and here set equal to 0 I should see on my calculator that it has zeros at 0 solving the second equation 3 over 2 in the third equation negative 1/2 again if I have time I will try to squeeze in that section and show you all my calculator how to solve it from 13 fifteen we’ve got solve X to the fourth plus four x squared minus 12 equals zero

this is not quadratic but it does have what we call quadratic form to it and for these problems you’ll find in our book that these things are always factorable x squared and x squared and to get 12 2 & 6 and I’ll make that positive and negative that will foil out back to what we started with 3 & 4 also give you 12 but 3 & 4 don’t have a difference of 4 in the middle so you have to use the 2 & 6 for your factorization now this is factored but it’s not factored down to linear binomials but that’s alright once I get it factored I can set each factor equal to 0 and then solve it any way that I know how to in this case I would do square rooting both sides square root of 2 is plus or minus root 2 x squared equals negative 6 square root of both sides okay I get plus or minus square root of a negative is I root 6 so there’s two answers plus or minus I root 6 I root 6 and negative I root 6 and root 2 and negative root 2 the problem 14 factor completely gave factor another factoring problem so obviously that’s gonna be important on the quiz and I probably shouldn’t have had it as y equals 0 I should have had a 0 instead of y there however I was trying to give you a clue on how to get started with factoring that because that thing has four terms and it’s cubic I’m not sure how to factor it however you can go to your calculator and you can graph that equation it’s X cubed and again I’ll try and show this on my calculator if at that time 2 x squared – 5 X and I’m just typing it in now to my calculator plus 6 and I’m gonna do for that a zoom 6 remember that gives us the best look at it and when i graph that right away i see it has a 0 at negative 1 don’t know what that is zero at negative 2 rather at 1 and it looks like at 3 and so if I’ve got my 0 so I can say it factors as X plus 2 X minus 1 and X minus 3 so there I’ve used my calculator to help me find the zeros or find the zeros which helps me find the factors okay lastly problem number 15 don’t have one so that’s all of the problems that I have for you to review plus the other 5 that we did in class so study practice the review guide that I gave you on Thursday that we went over on Friday I’ll try and get both of those things posted this video and the solutions to them good luck tomorrow