How do you convert polynomials into different bases?

Question

I'm wondering how you go about converting a polynomial, such as:

(x^3)+4(x^2)+3(x^1)+1

into a different base, say base (x-2)?

I guess I'm not being clear enough,

lets say we conver the number 16 (in base 10) into base 5...

16 = 1(10^1) + 6(10^0)

well in base 5, 6(10+0) = 1(5^1) + 1(5^0)
and 1(10^1) in base 5 = 2(5^1)+0(5^0)

so 16 in base five is 1(5^1)+1(5^0)+2(5^1)
or 3(5^1)+1(5^0)
ooooor

31 (base 5)

note: to check, we have 3-5's and 1-1. 3(5)s in base 10 is 1-10 and 5-1s (more commonly known as 15) + 1-1 is 1-10 and 6-1s (i.e., 16).

okay, maybe what you're saying is right. let's say that we're converting 7(x^2)+5(x)+4 into base (x-2)

we'll use (y-2)...

=7((y-2)^2) + 5(y-2) + 4
=7((y^2)-4y+4)+5y-2+4
=7(y^2)-28y+28+5y-2+4
=7(y^2)-23y+30

now i substitute for y = x-2... so

7 ((y-2)^2)-23(y-2)+30

Alright, i figured it out after a little bit of help... i'll post the result so you all can see.

we're going to convert 7(x^2) + 5x + 4 into base (x-2).

First we note that x = (x - 2) + 2.
so let y = x - 2 so (x^2) = (y +2)^2

so we have 7x^2+5x + 4=7((y+2)^2)+5(y+2) + 4.

which is equal to
7(y^2+4y+4) + 5(y+2) + 4
=
7y^2 + 28y+28+5y+20+4
=
7y^2+33y+52

subbing back in we get

7(x-2)^2+33(x-2)+52

Answer 1

Everywhere there is an x, put an (x-2) instead.

Expand terms as needed.
I would recommend using a different letter variable, say x=(y-2), for clarity.

(y-2)^3 + 4(y-2)^2 + 3(y-2) + 1...

Answer 2

I asked myself same question and one guy suggested me to use Taylor series,

So p(x) = x^3 + 4x^2 + 3x +1 = p(2) + (x-2)p'(2) + 1/2 (x-2)^2 p''(2) + 1/6 (x-2)^3 p'''(2), where p' is the derivative, p''' =1 actually.