# Write the polynomials in x as polynomials of

1. May 6, 2012

### JoshMaths

The problem statement

Write the following polynomials in x as polynomials of (x-3)

Solution should be somewhat analytical in its approach.
How would you do something like this? What does it mean?

You can use any example to explain it, my specific homework question isn't necessary unless you want to see it.
eg: a) x^4
b) x^2-x-4

The attempt at a solution
i am guessing it is not as simple as
x = x - 3 but some guidance would be great.

2. May 6, 2012

### Infinitum

As I see it, it means you have to represent the polynomial in the form of another variable y = x-3. For example, f(x) = x + 10 would be changed to f(y) = y+13, where y = x-3.

Though this sounds too simple, somehow.

3. May 6, 2012

### JoshMaths

Yes this is what i thought also, maybe I just lucked out with a really easy question ;)

4. May 6, 2012

### tiny-tim

Hi JoshMaths!

(try using the X2 button just above the Reply box )

Show us your answer for x2 - x - 4, just to be on the safe side.

5. May 6, 2012

### JoshMaths

Got it, thanks for the tip ;)

And fwi i am doing college level maths haha so algebriac manipulations aren't the pressing issue but to make you happy...

"x2 - 7x + 8"

6. May 6, 2012

### tiny-tim

hmm …

that's the trouble with using the same letter to mean two different things
x' = x - 3, and you need to convert from x to x'

(then you can rename it "x")

7. May 6, 2012

### epenguin

I would use the formula (which you can find explained early on - p.8 - in the book of Burnside and Panton* which I have mentioned earlier) which they state essentially as

f(x + h) = f(x) + f ' (x) h + f ''(x) h2/1.2. + f '''(x) h3/1.2.3 +... + f(n)hn/n!

You need to turn this around, you are trying to express in terms of a new variable x' = (x - 3). You have f(x); that equals f(x' + 3)
Develop that according to the above formula.

*(I commented this book in a thread called 'refer me to the best book in algebra' and I have just seen that it is available free online.)

This could be called "somewhat analytical in its approach" inasmuch as it uses derivatives. Although if you look harder you can see that you could redefine derivatives for polynomials in a finitistic purely algebraic way not involving limit concepts; they would be just the same formulae as the familiar ones. I think that may be what they have in mind by that phrase. If this is obscure not to worry - just do the calculation.

Last edited: May 6, 2012