1. last polynomial

P(x)=3.P(-x)+4x-1
Find remainder when P(x-1) is divided by (x+2).
:
My work:

P(-3)=3.P(3)-13

there is no relation between P(x) and P(-x),middle term of both equations are different.One has a - sign,the other has not..

2. Re: last polynomial

well.. one can always go the brute force route

$p(x) = 3p(-x) + 4x -1$

let $p(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 \dots$

$p(-x) = c_0 - c_1 x + c_2 x^2 - c_3 x^3 \dots$

$c_0 = 3 c_0 -1 \Rightarrow c_0 = \dfrac 1 2$

$c_1 = -3c_1 + 4 \Rightarrow c_1 =1$

$c_2 = 3 c_2 \Rightarrow c_2 = 0$ and similarly for $c_k,~k>2$

$p(x) = x + \dfrac 1 2$

$p(x-1) = (x-1) + \dfrac 1 2 = x - \dfrac 1 2$

$\dfrac{x-\frac 1 2}{x+2} = -\dfrac 5 2$

3. Re: last polynomial

How did you get 1 and 1/2,3c00[/S]?

4. Re: last polynomial

How did you find C_{0},C_{1},C_{2}?

5. Re: last polynomial

$$P(x)=3P(-x)+4x-1$$
$$P(-x)=3P(-(-x))+4(-x)-1 = 3P(x)-4x-1$$
Substitution yields
$$P(x)=3(3P(x)-4x-1)+4x-1$$
$$P(x)=9P(x)-12x-3+4x-1$$
$$-8P(x)=-8x-4$$
$$P(x)=x+\frac{1}{2}$$

6. Re: last polynomial

Ahri,Many Thanks. I undersood.But I am curious about thesolution above.

How did he find C_{0},C_{1},C_{2}?

7. Re: last polynomial

Many Thanks . Ok I understood the solution but how did you find c0,c1 and c2?

8. Re: last polynomial

By comparing coefficients of the constant term, $x$ term, $x^2$ term, and so on.

9. Re: last polynomial

Originally Posted by romsek
well.. one can always go the brute force route

$p(x) = 3p(-x) + 4x -1$

let $p(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 \dots$

$p(-x) = c_0 - c_1 x + c_2 x^2 - c_3 x^3 \dots$

$c_0 = 3 c_0 -1 \Rightarrow c_0 = \dfrac 1 2$

$c_1 = -3c_1 + 4 \Rightarrow c_1 =1$

$c_2 = 3 c_2 \Rightarrow c_2 = 0$ and similarly for $c_k,~k>2$

$p(x) = x + \dfrac 1 2$

$p(x-1) = (x-1) + \dfrac 1 2 = x - \dfrac 1 2$

$\dfrac{x-\frac 1 2}{x+2} = -\dfrac 5 2$
the last line should be

$x-\frac 1 2 \pmod{x+2} = -\dfrac 5 2$

10. Re: last polynomial

Originally Posted by romsek
the last line should be
$x-\frac 1 2 \pmod{x+2} = -\dfrac 5 2$
Reply #5 is more beautiful solution.