another polynomials question (1 Viewer)

[243_robbo]

another polynomials question

for a polynomial of say degree 4, let the roots be a, b, c and d. With the questions that ask you to find the polynomial with roots a/2, b/2, c/2, d/2 or a^2, b^2, c^2, d^2. i remember there are 2 ways of doing these types of questions, a short way and really long winded way. does anybody know any links to sites or previous posts that show both ways and compare them or something.

any help would be appreciated

 

[Trev]

Short way:
If you want to find the polynomial with the roots a/2, etc. Let x=a/2 so a=2x, and sub 2x into the equation. Similarly for x=a² sub in a=x^1/2.

 

[Riviet]

Let's say we have a quartic equation in p(x). If the old roots are a, b, c, d, and we want to find the polynomial with roots a/2, b/2, c/2, d/2 for example,

let x=a, b, c, d and y=a/2, b/2, c/2, d/2.

Then y=x/2
x=2y

Since x is a root of the polynomial, substitute x=2y into your original quartic equation:

p(2y)=0

In some cases, you will need to manipulate your new equation to get it into the general form of what your original equation was. For the quartic one, it would be Ax⁴+Bx³+Cx²+Dx+E=0

Hope that explains it.

 

[243_robbo]

wow you guys are so helpful, that post would've been up for what 10 minutes and ive already got the answer im looking for, thanks.


so in regards to the long way, ie using a + b + c + d = -b/a ; ab + ac + ad + bc + bd +cd =c/a ; etc.... results.


basically stuff em and use the above method?

or are there any kinds of questions where the long way is the only one taht will get an answer

(would this be comparable to differentiation and using first principles?)

 

[Riviet]

basically stuff em and use the above method?

or are there any kinds of questions where the long way is the only one taht will get an answer

(would this be comparable to differentiation and using first principles?)

Yes, and I haven't encountered a question that requires the long way, and in a way yes.

 

[pLuvia]

Sometimes they give you roots which are like

a+1/a , b+1/b, c+1/c, d+1/d

Then you have to find the new equation Then it might get tricky.

 

[c0okies]

243 robbo your picture in the signature space is really disturbing, could you please move it or make it smaller

 

[243_robbo]

is it too big, how much smaller would you liek it, to the nearest ratio divisible by e^6 and then express in polar for and perform a trig substitution on it while looking atthe screen upside down

 

[243_robbo]

happy?

 

[schismular]

wow you guys are so helpful, that post would've been up for what 10 minutes and ive already got the answer im looking for, thanks.


so in regards to the long way, ie using a + b + c + d = -b/a ; ab + ac + ad + bc + bd +cd =c/a ; etc.... results.


basically stuff em and use the above method?

or are there any kinds of questions where the long way is the only one taht will get an answer

Dont you have to use the longer method when evaluating- ie P(x) has roots a, b, c, d- find what a^2 + b^2 + c^2 + d^2 is equal to? So basically, if you have to evaluate what something is equal to (as above) tehn you use the relationships between the roots, whereas when actually finding the new equation with a new set of roots you use the shorter method.

 

[c0okies]

yea alot better; thanks

 

[Riviet]

Dont you have to use the longer method when evaluating- ie P(x) has roots a, b, c, d- find what a^2 + b^2 + c^2 + d^2 is equal to? So basically, if you have to evaluate what something is equal to (as above) tehn you use the relationships between the roots, whereas when actually finding the new equation with a new set of roots you use the shorter method.

Yep, you got the idea. =)