-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
Help me..factoring polynomials! (1 Viewer)
- Thread starter IAU001
- Start date
This is quite tricky - its especially hard to show a motivation for finding the solution but heres some ideas on how i got the factorisation.
Notice firstly that c is special variable in a sense that the expression is symmetric in a and b (ie you can swap them around and get exactly the same expression) but not in c.
So we then think of this expression as a function in c - i.e.:
f(c) = a^4 + b^4 +c^2 – 2 (a^2*b^2 + a^2*c + b^2*c) (I should learn how to do super- and sub- scripts, sorry)
now we know that the factor theorem states that if we sub in another value x for c and f(x) = 0 then (c - x) is a factor. Unfortunately, the coefficients/constants in this equation are not numbers but are other variables so x has to be some expression involving a and b.
We can make a couple more educated guesses about what x could be by noticing that every c in this expression has and exponent that is half that of the corresoponding exponent of a or b (ie c^2 instead of a^4 and b^2*c instead of b^2*a^2). Thus we assume that the degree of the expression we make x should be 2 (ie a quadratic).
So now its basically just trial and error so u might test c = a^2 + b^2 first (as a simple choice) and if that doesnt work, test maybe c = a^2 + ab + b^2 etc.
I know its a complicated (and not the best worded) explanation but its the best way I can provide a method for getting the answer without giving it to you (because I sense this is supposed to part of some enrichment programme because there is no way that this is yr 10 maths from the syllabus).
[Oh and Mark576's expression is not a complete factorisation because it is the sum of two terms.
Notice firstly that c is special variable in a sense that the expression is symmetric in a and b (ie you can swap them around and get exactly the same expression) but not in c.
So we then think of this expression as a function in c - i.e.:
f(c) = a^4 + b^4 +c^2 – 2 (a^2*b^2 + a^2*c + b^2*c) (I should learn how to do super- and sub- scripts, sorry)
now we know that the factor theorem states that if we sub in another value x for c and f(x) = 0 then (c - x) is a factor. Unfortunately, the coefficients/constants in this equation are not numbers but are other variables so x has to be some expression involving a and b.
We can make a couple more educated guesses about what x could be by noticing that every c in this expression has and exponent that is half that of the corresoponding exponent of a or b (ie c^2 instead of a^4 and b^2*c instead of b^2*a^2). Thus we assume that the degree of the expression we make x should be 2 (ie a quadratic).
So now its basically just trial and error so u might test c = a^2 + b^2 first (as a simple choice) and if that doesnt work, test maybe c = a^2 + ab + b^2 etc.
I know its a complicated (and not the best worded) explanation but its the best way I can provide a method for getting the answer without giving it to you (because I sense this is supposed to part of some enrichment programme because there is no way that this is yr 10 maths from the syllabus).
[Oh and Mark576's expression is not a complete factorisation because it is the sum of two terms.
Last edited:
IAU001
Member
- Joined
- Dec 23, 2006
- Messages
- 47
- Gender
- Male
- HSC
- 2009
I don't understand how you incorporated 'x' into the expression. I think I understand the part where you say that if f(x) = 0 then (c - x) is a factor (because (c - x) = c, right?). If my previous statement is correct, then I do not understand how x has to be some expression which involves a and b.AMorris said:
Sorry for being such a hindrance and thanks for your patience.
Umm - this is quite tricky to understand and even harder to explain over a forum but i think that if i just gave you the factorisation that would be quite useless so i will try to continue explaining.
lets start of with something easier first to explain the factor theorem.
let f(n) = n^2 - 5n + 6. now f(3) = 9 - 15 + 6 = 0. so by the factor theorem (n - 3) is a factor of f(n).
So now we move onto something a little trickier
let f(a) = a^2 - b^2.
if we substitute b for a in this expression we get f(b) = b^2 - b^2 = 0. So now we know that (a - b) is a factor of this expression.
For another example, let f(a) = a^2 + ab - a*b^2 - b^3. then f(b^2) = b^4 + b^3 - b^4 - b^3 = 0. so (a - b^2) is a factor of f(a).
(I'm pretty sure (~90%) the factor theorem can be extended in this way to polynomials in multiple variables but I could be corrected.)
so basically we are trying to find something to substitute for c in our expression such that it becomes equal to 0. this expression will most likely (we take a bit of an educated guess here - im not sure if this is necessarily true, but it is in this case) be a function of both a and b.
[What I've done in my previous explanation is let this expression be called x (which is a function in a and b - but thats also complicated to understand).]
We make a further educated guess that the expression x has each term of degree 2 (i.e. the powers of each term add to 2 e.g. a^2, ab, 5*a^3/b are all terms with degree 2.). And then we just guess things for x to be equal to, starting with simple stuff.
I hope this clarifies things, but I'm not sure if it will.
lets start of with something easier first to explain the factor theorem.
let f(n) = n^2 - 5n + 6. now f(3) = 9 - 15 + 6 = 0. so by the factor theorem (n - 3) is a factor of f(n).
So now we move onto something a little trickier
let f(a) = a^2 - b^2.
if we substitute b for a in this expression we get f(b) = b^2 - b^2 = 0. So now we know that (a - b) is a factor of this expression.
For another example, let f(a) = a^2 + ab - a*b^2 - b^3. then f(b^2) = b^4 + b^3 - b^4 - b^3 = 0. so (a - b^2) is a factor of f(a).
(I'm pretty sure (~90%) the factor theorem can be extended in this way to polynomials in multiple variables but I could be corrected.)
so basically we are trying to find something to substitute for c in our expression such that it becomes equal to 0. this expression will most likely (we take a bit of an educated guess here - im not sure if this is necessarily true, but it is in this case) be a function of both a and b.
[What I've done in my previous explanation is let this expression be called x (which is a function in a and b - but thats also complicated to understand).]
We make a further educated guess that the expression x has each term of degree 2 (i.e. the powers of each term add to 2 e.g. a^2, ab, 5*a^3/b are all terms with degree 2.). And then we just guess things for x to be equal to, starting with simple stuff.
I hope this clarifies things, but I'm not sure if it will.
IAU001
Member
- Joined
- Dec 23, 2006
- Messages
- 47
- Gender
- Male
- HSC
- 2009
I understand everything in your post except for the bold part in your quote. Is that a typo? Because 5a<sup>3</sup>/b would be of degree 3 right? Since 5 is of degree 1 and 'a' is of degree 3, which we add makes degree 4; 'b' has a degree 1, which we then subtract from 4 to make an answer which should be of degree 3. If I am correct in saying this, then yes I understand the whole thing in that post.AMorris said:
Is trial and error the only way? Does that mean that any similar questions that I get like this (assuming that it comes in Westpac Maths Comp or wherever) I have to use trial and error? Or is there another miracle method which you can pull for me from thin airAMorris said:
.Thanks a lot by the way. This is a series of amazing revelations for me, being in year 10.
Oh excellent. Just clarifying the bold bit: 5 actually has degree 0 (just like how the constant polynomial has degree 0) Possibly one way to think about this is that 5a^3/b could be written a^3/b + a^3/b + a^3/b + a^3/b + a^3/b. and now its clear that each term has degree 2.
I don't have any better method for approaching it other than trial and error but you can make educated guesses based on what you want in your final thing. like since u want an a^4 term you can get that most simply by multiplying a^2 * a^2 (or a^3 * a) so u guess that one of those is in the answer. I have no other special method for determining those factors beyond that and if it did have a really ugly factorisation then i'd be pretty stuck.
(Another thing that's quite interesting that I've observed about this expression is that if u replace c by d^2 then the expression becomes completely symmetric in a, b, d. However this is in a way not allowed to be done because c has to be positive (or 0) if there is to be a real d such that d^2 = c. But thats just an aside)
I don't have any better method for approaching it other than trial and error but you can make educated guesses based on what you want in your final thing. like since u want an a^4 term you can get that most simply by multiplying a^2 * a^2 (or a^3 * a) so u guess that one of those is in the answer. I have no other special method for determining those factors beyond that and if it did have a really ugly factorisation then i'd be pretty stuck.
(Another thing that's quite interesting that I've observed about this expression is that if u replace c by d^2 then the expression becomes completely symmetric in a, b, d. However this is in a way not allowed to be done because c has to be positive (or 0) if there is to be a real d such that d^2 = c. But thats just an aside)