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HLC: Algebraic long division problem
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13-02-2012 4:31pm#1
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I'm not sure I fully understand the problem you are having, are you saying that in these type of problem you are having trouble figuring out the (a-2) part needed to compete the problem?
Yes. It seems sometimes I choose a second term that will allow me to cancel what's attached to x^2, other times what's attached to x, and other times the constant. When I say cancel I mean negate; like putting -(a-2)x^2 under (a-2)x^2.Maybe you could post another example of where you are having trouble and I will try to help as best I can.*
I am stuck on this problem again but bear in mind the problem may require the factor theorem which is covered later in this section, the reason I'm attempting it is because it's part of the revision questions.
That said, it does look identical to many of the other exercises that do not require the factor theorem - all the more confusing unfortunately.
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If (2x - 1) divides evenly into the cubic then x = 1/2 makes the whole expression equal to zero. [Since (2x-1) is a factor, it becomes zero when x = 1/2).
Now if you sub in the value x = 1/2 into the original cubic expression and put the result equal to zero, you get k = -13.
So, go back to the original question with this value of k, do the division and get the other (quadratic) factor to be x^2 + 7x + 12, which factors nicely into (x + 4)(x + 3).
[Note that these two factors mean that the other 'roots' of the original cubic are -4 and -3.];)0 -
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If (2x - 1) divides evenly into the cubic then x = 1/2 makes the whole expression equal to zero. [Since (2x-1) is a factor, it becomes zero when x = 1/2).
Am I correct that you chose to make x = 1/2 because that is what required to make the factor equal to zero? So if say the factor was (2x-4) we would instead make x=2?Now if you sub in the value x = 1/2 into the original cubic expression and put the result equal to zero, you get k = 13.
I understand what you have done so far and I am able to get 13 out of the equation but I do not understand why we make the factor equal to zero in the first place but if I did I think it would help a lot. Perhaps you could rephrase?So, go back to the original question with this value of k, do the division and get the other (quadratic) factor to be x^2 + 7x + 12, which factors nicely into (x + 4)(x + 3).
That I follow![Note that these two factors mean that the other 'roots' of the original cubic are -4 and -3.];)
That too, thanks!I'm not 100% sure on this but I'm pretty confident. To solve this you need to look at the last two terms i.e 17x-12......Now for to solve this we need something that is divisible by 2x-1 thus we need to increase the 17x to 24x when finding the 2nd term.
How do you know to increase 17x by exactly 7x to give 24x? Why not just increase to 18x which is also divisble by 2x?This means that the 2nd term would be 7x giving k+1=14 and then leaving us with 24x-12. This means our final term is 12 leaving us with a quadratic of x^2+7x+12.
I am sorry, I do not understand this method
(but I would like to understand it!) 0 -
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I have now come onto the part of the notes I am learning from that explains using the factor theorem in this way, i.e. solving the factor = 0.
I understand that to prove a binomial is a factor of an equation I need to solve the factor for x, and then substitute the value for f(x).
An example here from my notes is this:
Use the Factor Theorem to show that (x-3) is a factor of x^3 - 5x^2 + 11x - 15.
Solution
Let f(x) = x^3 - 5x^2 + 11x - 15
Putting x-3 = 0, we get x=3. Thus we have to show that f(x) = 0.
f(3) = (3)^3 - 5(3)^2 + 11(3) - 15
= 27 - 45 + 33 - 15
= 0
Thus (x-3) is a factor of f(x).
I was curious to see if I could also still use the old method to prove a factor correct by dividing it into the cubic and determining a remainder of 0.
This worked find giving a remainder of 0 and x^2 - 2x + 5, but when I tried to factor that it could not be done easily by the quick method but even when using the factor theorem it would not work out because it gave a minus square root?
I know the value of the other factors were not asked in the question but I should still be able to derive them, no?0 -
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If its not asked in the question, I'd say forget about it altogether. Having said that however it should still be possible to work out the other factors for most problems. If you have covered complex numbers this is actually fairly simple to solve. I wont go into it now as I dont know if you have complex numbers covered and if you dont it wont make any sense.
Ah I remember hearing something before about negative square roots and imaginary (i) numbes but you're right it is something I have still to come to so I will wait for it
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Thanks for that Mwalimu.
How does one decide, when given a factor and a polynomial with a hidden coefficient and asked to find the hidden coefficient and the other two factors which method of the two following to use?
Put the factor equal to zero to find the root, then put the polynomial with the found roots visible equal to zero to find the hidden coefficient before dividing the factor into the polynomial with the coefficient visible to find the remaining factors.
Use long division equating like with like to first find the coefficient, and then divide the factor into the polynomial with the coefficient visible to find the remaining factors.
I ask because the notes I am working for shows the solution for each exercise and in this more recent section of 15 questions or so I have been using the factor theorem method because that is what the last examples did, but now on the fourth question when checking my answer the solution for it has in brackets (Division is probably the better option here).
The same answer is found, I am just unsure why one method is recommended over another? Also, should I be able to use both methods all the time?
I have included my work for the question using both methods respectively.
The text missing top right from the image above is:2x - 1 = 0 2x = 1 x = 1/2
I see the second method is shorter, but how will I know when to use it?
To see if I could use both methods I tried method 2 on an earlier exercise I solved using method 1 but it did not seem to work:
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Firstly sorry for the late reply, I went on studying the method of guessing the missing root based on the factors of the constant and I have also been watching some Eircom Study Hub videos.If you had continued with your final example by doing the long division, it would work out fine. The next term would be (k+2)x so that you can eliminate the (k+2)x^2.
Forgive me, but how do you know it is the (k+2)^2 you have to elimite. It is clear that sometimes we elimite the x^2 term, sometimes the x term, and at other times the constant as you will see in a question I will come onto.When you do the multiplying and subtracting, you get the following remainder: (18 + 2k)x -14 which means the final part has to be (18 + 2k) in order to eliminate the x term. After dividing, the remainder (when you tidy it up) is then 20 + 4k. However, since (x - 2) is a 'factor' of the original expression, the remainder should be zero. Hence you find k = -5 which is exactly what you get if you sub in x = 2 [from letting x - 2 = 0] to the original expression. You also get 4k + 20 and then putting that equal to zero gives k = -5.
I got that far, please see my work below, but I got stuck further down trying to solve the equation when forced to use the factor theorem discovering it was trying to square root a negative (-23).
Hope all this helps. Keep the maths going strong!;)
It does help, and is very much appreciated
The next question:
x^2 + 5px + pq is a factor of f(x) = x^3 + qx^2 - (p^2)x - (p^2)q.
Write all solutions to the equation f(x) = 0 in terms of p.
Visible in the screenshot below on my first attempt I got stuck on the fourth line of the long divsion where -p was required to eliminate the constant so I had to look at the solution. How does one know that this time it's the constant needing to be eliminated?
For practice I then tried to use the method of creating a factor (x+k) and then finding out what k would be equal to and then q but I did not see q = 4p coming out of the equation - did I make a mistake somewhere?
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I just want to upload this problem while there are people online answering, if the information I seek is answered in your most recent replies you need not address it.
The instructor from study hub/exam support uses the adding an (x+k) factor method which I will do but I would like to always be able to solve using the long division method also.
I only put 3 as the second term of the quotient because it made b = a^2, was I correct to do so? What I am asked to find is top right of the page.
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MathsManiac wrote: »Your first 5 lines are correct, but then when you equate coefficients to pull out the three equations, you've made a mistake on the middle and last ones. The RHS of the middle one should be -p^2 and the RHS of the last one should be -p^2q.When you got to the two cubics being put equal to each other and compared terms, you should have got 5kp + pq = -p^2 for the x coefficients (you forgot the square on the 'p'), and kpq = -p^2q for the constant term (you forgot to include the final q).
This last part then simplifies to give k = -p which you can use with the x^2 coefficients [k + 5p = q] to yield 4p = q, as was the case originally.
I will get to this right after posting!MathsManiac wrote: »Since the constant term in your quadratic factor was pq, and you needed to finish up with a constant term of -p^2q in the cubic, the constant termin the linear factor has to be -p. So, you could have just multiplied (x-p) by (x^2 + 5px + pq) and equated coefficients. This obviates the need to introduce another letter (k).
I am going to try that method too!
Further to what you have said could I remember it thus; whenever the constant term in the divisor (quadratic and linear?) shares an unknown(or variable? *not sure of exact nomenclature*) with the constant of the cubic (or quadratic?) to be divided into, we can use that unknown with the sign of the unknown in the expression we are dividing into as the second term of the factor we create?1) Forgive me, but how do you know it is the (k+2)^2 you have to elimite. It is clear that sometimes we elimite the x^2 term, sometimes the x term, and at other times the constant as you will see in a question I will come onto.
Since the first term in the remainder after you have eliminated the x^3 is the x^ term, that's the one to focus on next. It's coefficient in this case is (k + 2). To eliminate it, you need to multiply the (x - 2) by ((k + 2)x).
That makes sense to me, and seems natural but I question it and hesitate because I have noticed on some of the algebraic long division questions when we are dividing into a cubic with one or two unknowns (a,b,p,q etc) we seem to do what does not seem natural to me when we are 'insisting' the remainder is 0. This is possibly a bad question to have asked how do you know, but I will link a question like I am talking about and one where I followed the solution according to the IoE notes at the bottom of this post.2) I got that far, please see my work below, but I got stuck further down trying to solve the equation when forced to use the factor theorem discovering it was trying to square root a negative (-23).
Depending on what you were asked to do originally (find k?), you may not need to solve the quadratic to get all the roots. If you do need to get all the roots, then this is an example where the roots of the original cubic involve complex numbers.
I don't think I was asked to find the roots but I was just doing it for practice.MathsManiac wrote: »As Mwalimu indicated in his last post: when doing long division, at each stage you are always trying to get rid of the highest remaining power, by seeing how many times the highest power in the divisor goes into the highest remining power in the latest remainder.
Yes, that makes sense and I have no problem with regular non-algebraic long division, and I now have no problem with regular alebraic long division. It's when there are unknown coefficients, especially two, with division that seems unnatural to me that I run into difficulty.
I will link a question like I am talking about and one where I followed the solution according to the IOE notes at the bottom of this post.MathsManiac wrote: »In your latest example you haven't followed this procedure correctly at the second stage.
You started correctly by basically saying "2x^2 into 2x^3 goes x times", so you put the x on top. Then you multiplied the x by the entire divisor, and subtract it off.
The next stage should then have been: "2x^2 into 6ax^2 goes 3a times", so you (should have) put 3a on top (not just 3) before proceeding.
I did have 3a there initially because that seems natural to me but I tipexed out the a when I was not getting b = a^2 further down!
As I have said it seems sometimes when there are unknown co-efficients we force a term which may appear not to make sense just to cancel something on a line when insisting the remainder is 0, so I thought that's what I was doing then despite my gut telling me 3a was correct as it's easily seen that's what 6ax^2 / 2x^2 is!
I will link a question like I am talking about and one where I followed the solution according to the IOE notes at the bottom of this post.MathsManiac wrote: »Are you sure that you've taken this question down correctly? As you have presented it, it doesn't work out. (That is, you are being asked to prove something that is demonstrably false - so you're on a hiding to nothing!) Either you've taken it down wrongly, or there's an error in the question.
I'm assuming from your work that the question reads like this: Given that 2x^2 - 3ax + b is a factor of 2x^3 + 3ax^2 - cx +3a^2, show that c = 8a^2 and b = a^2.
That is a false statement (by which I mean that I can provide a counterexample) and therefore you have no hope of proving it.
From the PDF:
From the video:
His solution:
I will link a question like I am talking about and one where I followed the solution according to the IOE notes at the bottom of this post.
I need only go back as far as question 110 which has already been mentioned on this thread.
Please direct your attention to the fourth line, where to me this does not seem like natural division, instead it is described in my notes as 'insisting' the remainder is 0.
If I were to do this the same as other long division mentioned I would look at the first term of the fourth line which is (-5p+q)x^2 and I would say that x^2 divided into that gives (-5p+q) and I would write that at the top for the second term of the quotient.
Instead according to the IoE solution notes a x^2 divided into the fourth line somehow gives a -p! I do see how this makes it so the constants cancel and we can equate the x^2 and x coefficients, but how does one know to not only skip the x^2 term, and the x term to cancel the coefficient when in what I call regular algebraic division it's the highest x power we need to cancel but I have also seen some solutions where it's neither the x^2 or constant term cancelled but instead the x term!
And how does one know that it's -p that needs to be written up top? I mean I guess when you know what needs to be cancelled you can get it easily, but this does not even seen like division any more because dividing x^2 in does not give -p!
Can you see where I am differentiating the different algebraic long division from natural and what seems unnatural? By unnatural I mean skipping dividing the highest power of x to do something else.
This is why on the earlier question I tipexed out the a from the 3a term in the quotient.
I know there's a lot in my posts and a hell of a lot of questions, I really appreciate the help you guys are giving me, thank you.0 -
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MathsManiac wrote: »You may have noticed (or maybe you didn't) that I was right about there being an error in the question. From the screenshot of the video, you will see that the second line of the solution indicates that the cubic was meant to end in 3a^3, not 3a^2.
I only really looked at the solution video for a second yesterday when I was taking the screenshot for my last post so I did not notice! I was working from the PDF file because I did not want to see his solution, although I suspected he would use the make up a factor method because he recommends it in the eLesson video.
I have been away from home since yesterday evening so I have only corrected my work with the different solutions and methods advised to me in this thread the night before but I left question 3 of the exam support module incomplete.MathsManiac wrote: »That last scanned page indeed presents a very unnatural way to do things - it's like starting out with a long division method, and switching to another halfway through.MathsManiac wrote: »It looks to me as though the person presenting this made a sort of end-run around the last bit of the division, and used what had to happen with the last term to simplify the remaining work. In that sense, it's a mixture of the regular long division method and the "create the other factor" method.
I know, and my not grasping just this part of the algebraic long division has been driving me crazy!
The author of the notes is Aidan Roantree, he teaches at the Institute of Education and has a mathematics book together, but I find no tutorial in either outlining the switch of tactic, or how to approach it etc.MathsManiac wrote: »You could of course instead have continued with long division "properly", by putting (q-5p) on top, and multiply it down. When you then set the remainder equal to 0, you get:- 5p(q-5p) = -p^2-pq, and
- p^2q = pq(q-5p)
Putting (q-5p) does make more sense to me and I'm going to do it all over again like that when I'm home in a few hours!MathsManiac wrote: »I certainly think that the quickest way of doing that question would have been to examine the constant coefficients at the start, which would tell you straight away that the linear factor is (x-p). Then, you could multiply (x-p) by (x^2 + 5pq +pq), and equate the coefficients to finish.
Yes I agree, I did it like that yesterday and it worked beautifully I just want the ability to approach any question with any working method which is why I'm still doing the long division on it.
I'm also going to look at doing the question that began this thread using the natural division method.0 -
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Please direct your attention to the fourth line, where to me this does not seem like natural division, instead it is described in my notes as 'insisting' the remainder is 0.
If I were to do this the same as other long division mentioned I would look at the first term of the fourth line which is (-5p+q)x^2 and I would say that x^2 divided into that gives (-5p+q) and I would write that at the top for the second term of the quotient.
Instead according to the IoE solution notes a x^2 divided into the fourth line somehow gives a -p! I do see how this makes it so the constants cancel and we can equate the x^2 and x coefficients, but how does one know to not only skip the x^2 term, and the x term to cancel the coefficient when in what I call regular algebraic division it's the highest x power we need to cancel but I have also seen some solutions where it's neither the x^2 or constant term cancelled but instead the x term!
And how does one know that it's -p that needs to be written up top? I mean I guess when you know what needs to be cancelled you can get it easily, but this does not even seen like division any more because dividing x^2 in does not give -p!
Can you see where I am differentiating the different algebraic long division from natural and what seems unnatural? By unnatural I mean skipping dividing the highest power of x to do something else.
This is why on the earlier question I tipexed out the a from the 3a term in the quotient.
I know there's a lot in my posts and a hell of a lot of questions, I really appreciate the help you guys are giving me, thank you.
Yes, this is what you do if following the routine method. However, after a while, it pays to take a broader look at what's going on and be a little smarter about what you do next. After you would have done the (-5p + q) multiplication out, you know that the x^2 term, the x term and the constant term are each going to have to equal zero. [Maybe this is what you mean by 'insisting' the remainder is zero?] Anyway, of the three possibilities, the constant gives the easiest way out. Looking to the constant term, whatever you get is going to have to equal -p^2q. Thus, the pq would have to be multiplied by -p [which is much neater to multiply by than (-5p + q)]. You seem to have realised that much in your own figuring out of what's going on. That's progress.
It just takes a little while for the confidence to come, so you can begin to look at the whole problem and make sensible decisions, based on a good understanding of the procedure. Following the rote procedure can get complicated at times, particularly if you don't really 'get it'. However, you seem to be more insightful and able to make some sense from the 'unnatural' ways of doing long division.
MathsManiac shows how the whole problem can be simplified by doing the 'overview' approach from the start, knowing what has to happen at the end.
As regards the other problem form the online support site,it's a pity the original question was had that error. It really is annoying when something as basic at that is wrong and throws the whole thing into confusion.:mad: However, look at the bright side - you have to really work though the problem to realise something is wrong and, by doing so, get a better understanding of the whole thing (although this was not the intention of the online lesson).0 -
MathsManiac wrote: »You may have noticed (or maybe you didn't) that I was right about there being an error in the question. From the screenshot of the video, you will see that the second line of the solution indicates that the cubic was meant to end in 3a^3, not 3a^2.As regards the other problem form the online support site,it's a pity the original question was had that error. It really is annoying when something as basic at that is wrong and throws the whole thing into confusion.:mad: However, look at the bright side - you have to really work though the problem to realise something is wrong and, by doing so, get a better understanding of the whole thing (although this was not the intention of the online lesson).
Worked out fine for me once I had the expression correctly. I think it's pretty sloppy of them not to correct mistakes like that when people are paying for a service! I of course am not paying and am accessing the study hub through my girlfriend's Eircom broadband... but stillMathsManiac wrote: »YSince the constant term in your quadratic factor was pq, and you needed to finish up with a constant term of -p^2q in the cubic, the constant termin the linear factor has to be -p.MathsManiac wrote: »The constant terms in the factors have to multiply to give the constant term in the cubic.
Equivalently, divide the constant term in the cubic by the constant term in one factor to get the constant term in the other factor.MathsManiac shows how the whole problem can be simplified by doing the 'overview' approach from the start, knowing what has to happen at the end.
Okay, I see that now and I am eager to come onto another question to see if I can spot and apply that method without prompting.MathsManiac wrote: »You could of course instead have continued with long division "properly", by putting (q-5p) on top, and multiply it down. When you then set the remainder equal to 0, you get:- 5p(q-5p) = -p^2-pq, and
- p^2q = pq(q-5p)
Got this out on paper just fine.Yes, this is what you do if following the routine method. However, after a while, it pays to take a broader look at what's going on and be a little smarter about what you do next.
Okay, I will bear that in mind.After you would have done the (-5p + q) multiplication out, you know that the x^2 term, the x term and the constant term are each going to have to equal zero. [Maybe this is what you mean by 'insisting' the remainder is zero?]
Yes, this is what is meant by insisting the remainder is zero from my notes.Anyway, of the three possibilities, the constant gives the easiest way out. Looking to the constant term, whatever you get is going to have to equal -p^2q. Thus, the pq would have to be multiplied by -p [which is much neater to multiply by than (-5p + q)].
It is neater, once I can determine the correct term to multiply. I have three more Examsupport exam model questions to work through, and 13 revision questions on this chapter from my notes, so I will try several methods on all of them.
In two questions mentioned earlier I am curious about the create a temporary factor method, (x+k) etc.
I had initially thought it would just be (x + k) I use but I encountered a solution using instead (ax + k), and on the next question I came onto there were some a coefficients so I tried (ax + k) but it did not work because I needed (x + k) this time.
How do we know which to apply?
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For this problem I have been given a quadratic factor and cubic product, and I have been asked to find the value of a and b.
I have gone over it but cannot see where I err as to not get the values listed at the back of the notes.
The notes are typed but the text is too small for my phone's camera to capture so I have transcribed them.
Please note I do the long division slightly different to the method outlined in the notes, but all I really change is the sign, and I believe my method closer to standard non-algebraic long division.
I am suspicious of the (a + 1) term changing to (a - 1) on the fifth line, can this be correct? Please see below.
To test to see if a different method confirmed either of the two previous I tried:
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You can usually check easily enough which answer is correct.
If you look at the notes' answer;
look at the two factors: (x^2 + ax + (a+1)) and (2x +1)
If you substitute in the notes' answer for a you get 2x^3 + 7x^2 +6x -5.
If you substitute in the notes' answers for a+b into the product you get 2x^3 +7x^2 +2x - 3
These do not match, but if you do the same for your answer, both will match, so just a slip of the pen from the note-maker. 0 -
BeanbagBallbag wrote: »You can usually check easily enough which answer is correct.
If you look at the notes' answer;
look at the two factors: (x^2 + ax + (a+1)) and (2x +1)
If you substitute in the notes' answer for a you get 2x^3 + 7x^2 +6x -5.
If you substitute in the notes' answers for a+b into the product you get 2x^3 +7x^2 +2x - 3
These do not match, but if you do the same for your answer, both will match, so just a slip of the pen from the note-maker.
Thanks! But I got different answers in the two different methods I tried?0 -
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In this next question I am given f(x) = x^3 + ax^2 + bx + c, that f(1) = 0, f(3) = 0, and f(2) = -f(0).
The solution given at the back of the notes lets the third factor be equal to (x + k) and then equates the three factors of f(2) to -[(0 - 1)(0 - 3)(0 + k)] to find k which is then used to multiply all the factors to show the cubic with visible coefficients.
a = -3, b = -1, c = 3
Should the different method I tried below not also work too?
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