How to do this (2 Viewers)

[ngai]

if A is a root and B is a root. Then AB is also a solution. as in, AB divides into P(x). Hence any product of two roots works with P(x).

what rubbish
consider the polynomial P(x) = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6
x = 2 is a root
x = 3 is a root
therefore x = 6 is a root, as 2*3=6
therefore, P(x), a cubic, has roots 1, 2, 3, 6
but also, x = 2 is a root, and x = 6 is a root
therefore x = 12 is also a root
therefore the same cubic, P(x), has roots, 1,2,3,6,12
and hence by an induction-ish thing, P(x) has infinitely many roots

 

[Logix]

right u r ngai well put

 

[Logix]

right u r ngai well put

 

[coca cola]

nobody's perfect

Hah thats true.

Well thanks a lot for the help guys.

 

[coca cola]

Is Geha's book anygood? Is it just worked solutions from past (how many years??) HSC papers grouped into topic by topic ? Inform me some1 thnx

Its a good book. Basically its just a problem type book where he groups specific kind of problems into one Tip and gives HSC examples on the top from past years and ratings of its difficulty. Then he gives his own examples and solutions which is somewhat harder than the hsc examples, some of the questions are just so long algebraicly that it takes you 3 or 4 lines to get a step of working out done. Its funny how he just states that it should be done that way and here is the solution.

 

[Jase]

what rubbish
consider the polynomial P(x) = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6
x = 2 is a root
x = 3 is a root
therefore x = 6 is a root, as 2*3=6
therefore, P(x), a cubic, has roots 1, 2, 3, 6
but also, x = 2 is a root, and x = 6 is a root
therefore x = 12 is also a root
therefore the same cubic, P(x), has roots, 1,2,3,6,12
and hence by an induction-ish thing, P(x) has infinitely many roots

lol, yes thats a problem.
But what i meant was.. if P(x) = (x-1)(x-2)(x-3)
then (x-2)(x-3) divides into P(x) perfectly. So you sub in the product of those roots randomly just to make it work for laziness and simplicity's sake.
What im saying is that the book is probably right, so if you get that in the exam just do it that way and if you get it wrong sue Jeff Geha! ^^

 

[mojako]

lol, yes thats a problem.
But what i meant was.. if P(x) = (x-1)(x-2)(x-3)
then (x-2)(x-3) divides into P(x) perfectly. So you sub in the product of those roots randomly just to make it work for laziness and simplicity's sake.
What im saying is that the book is probably right, so if you get that in the exam just do it that way and if you get it wrong sue Jeff Geha! ^^

it wont work ^_^
"then (x-2)(x-3) divides into P(x) perfectly" : true
"So you sub in the product of those roots randomly" not true

you need to sub in (2-2)(3-3), i.e.
you need to sub in two x values, one is 2 and the other is 3.
the third one is.. whatever you want..
actually any two of the three can be whatever you want.

what rubbish
consider the polynomial P(x) = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6
x = 2 is a root
x = 3 is a root
therefore x = 6 is a root, as 2*3=6
therefore, P(x), a cubic, has roots 1, 2, 3, 6
but also, x = 2 is a root, and x = 6 is a root
therefore x = 12 is also a root
therefore the same cubic, P(x), has roots, 1,2,3,6,12
and hence by an induction-ish thing, P(x) has infinitely many roots

Therefore, P(x) is a zero polynomial, and
a = b = c = d = 0
so Jase's statement is somewhat valid,
and not a complete rubbish.

 

[Xayma]

If A is a root and B is a root then (x-A)(x-B) is a factor of P(x). Not necessairly (x-AB).

 

[ngai]

What im saying is that the book is probably right, so if you get that in the exam just do it that way and if you get it wrong sue Jeff Geha! ^^

hmm...yeh, i understand ur point
well ur not from my school...so less marks for u means better results for us

Therefore, P(x) is a zero polynomial

i thought i said P(x) = x^3 - 6x^2 + 11x - 6?

 

[mojako]

i thought i said P(x) = x^3 - 6x^2 + 11x - 6?

oh yes, haha... stupid me
but.. we can change the definition of the symbols "6" and "11".
6 = 0 and 11 = 0 ; )

 

[Xayma]

It is a possible root, due to it being a factor of the x⁰ term but you are better off staying away from it.

Anyway if you posses x=2,x=3 and its a cubic and the last term is has a magnitude of 6 your other root will be 6/(2*3)=1 since they must times together to make the constant.

 

[CrashOveride]

it wont work ^_^
"then (x-2)(x-3) divides into P(x) perfectly" : true
"So you sub in the product of those roots randomly" not true

you need to sub in (2-2)(3-3), i.e.
you need to sub in two x values, one is 2 and the other is 3.
the third one is.. whatever you want..
actually any two of the three can be whatever you want.


Therefore, P(x) is a zero polynomial, and
a = b = c = d = 0
so Jase's statement is somewhat valid,
and not a complete rubbish.

Kinda off track but..
I think mojo is making the comment that if some polynomial vanishes for more than n values of x (where n is the degree >=1), it must be the zero polynomial.

 

[mojako]

It is a possible root, due to it being a factor of the x⁰ term but you are better off staying away from it.

Anyway if you posses x=2,x=3 and its a cubic and the last term is has a magnitude of 6 your other root will be 6/(2*3)=1 since they must times together to make the constant.

Hmm.. so
1. the product of roots must be +6 no matter whether the constant is +6 or -6
2. if the leading term doesn't have a magnitude of 1, the product of roots will still be +6
right?

The HSC syllabus tells you a different thing.
Stupid HSC

 

[coca cola]

Does any one know how I can contact Geha?

 

[mojako]

Does any one know how I can contact Geha?

Doesn't the book have a contact number?

Hmm.. you might be able to ask from this:
http://www.boredofstudies.org/jeffgeha.php

 

[ngai]

It is a possible root, due to it being a factor of the x⁰ term but you are better off staying away from it.

being a factor of the x0 doesnt show anything
u can have roots x=4235, x = 6/4235, x = 1, which would satisfy the x0 term being 6
and same with 23149878291347, 6/23149878291347, 1
and even (5234+9128347i), 6/(5234+9128347i), 1

 

[mojako]

being a factor of the x0 doesnt show anything
u can have roots x=4235, x = 6/4235, x = 1, which would satisfy the x0 term being 6
and same with 23149878291347, 6/23149878291347, 1
and even (5234+9128347i), 6/(5234+9128347i), 1

the numerator of any integer root of a polynomial with real coefficients will be a divisor of the constant term
and given an unknown polynomial, there is a non-zero probability that the polynomial has only real coefficients, and has an integer root, and that the leading term is 1 or -1

 

[Xayma]

But assuming you already now two of them. Using the products of roots αβγ=-d/a

If you know 2 and 3 it will work.

You normally try the integer fractions first because they have a good chance of being it.

Also those other examples would distort the other coefficients enough it would be obvious.