1. polynomials

Find a and b (from reals) so that

$\displaystyle p(x) = ax^{2008} + bx^{2007} - 1$
will be dividable by polynomial $\displaystyle (x - 1)^2$.

no idea how to begin.. can someone help?

2. Originally Posted by metlx
Find a and b (from reals) so that

$\displaystyle p(x) = ax^{2008} + bx^{2007} - 1$
will be dividable by polynomial $\displaystyle (x - 1)^2$.

no idea how to begin.. can someone help?
i think $\displaystyle P$ will be dividable by $\displaystyle (x-1)^2$ if $\displaystyle P(1) =0.$

3. Originally Posted by Raoh
i think $\displaystyle P$ will be dividable by $\displaystyle (x-1)^2$ if $\displaystyle P(1) =0.$
Yes, but I believe that this is only half of the story. 1 is a double root.

4. so how do i solve this?

I did this:

If $\displaystyle p(1) = 0 \Rightarrow a + b = 1$
$\displaystyle a = 1 - b$

$\displaystyle x_{1,2} = \frac{-b \pm \sqrt{b^2 -(- 4(1 - b))}}{2(1 - b)} = \frac{-b \pm \sqrt{b^2 - 4b + 4}}{2 - 2b}$

$\displaystyle = \frac {-b \pm \sqrt{(b - 2)^2}}{2 - 2b} = \frac{-b \pm (b - 2)}{2 - 2b}$

$\displaystyle x_1 = \frac{-b + b - 2}{2 - 2b} = - \frac {1}{1 - b} = - \frac {1}{a}$

$\displaystyle x_2 = \frac{ -b - b + 2}{2 - 2b} = 1$

does this help me in anyway?

i can put those values instead of x-es..
$\displaystyle a \cdot (-\frac{1}{a})^{2008} + b \cdot (- \frac{1}{1 - b})^{2007} - 1 = 0$

$\displaystyle a \cdot \frac{1}{a} - \frac{b}{1 - b} - 1 = 0 \Rightarrow -\frac{b}{1 - b} = 0 \Rightarrow b = 0$

a = 1 - b = 1.

is that how you do it?

5. Originally Posted by VonNemo19
Yes, but I believe that this is only half of the story. 1 is a double root.
try to use the synthetic division (Synthetic division - Wikipedia, the free encyclopedia) to divide $\displaystyle p(x)$ by $\displaystyle x^2 - 2x + 1$

hope it'll work

6. Originally Posted by dedust
try to use the synthetic division (Synthetic division - Wikipedia, the free encyclopedia) to divide $\displaystyle p(x)$ by $\displaystyle x^2 - 2x + 1$

hope it'll work
is that the same as Horner's algoritm?

7. Originally Posted by VonNemo19
Yes, but I believe that this is only half of the story. 1 is a double root.
exactly.

8. Originally Posted by metlx
is that the same as Horner's algoritm?
yes, with a slight modification

9. Originally Posted by dedust
try to use the synthetic division (Synthetic division - Wikipedia, the free encyclopedia) to divide $\displaystyle p(x)$ by $\displaystyle x^2 - 2x + 1$

hope it'll work
the degree of P is a LOT bigger than $\displaystyle x^2 - 2x + 1$,are you really sure about doing the division.

10. Originally Posted by Raoh
exactly.
so did i do it right or not? :P

11. Edit* Made some silly errors...

12. Originally Posted by Raoh
the degree of P is a LOT bigger than $\displaystyle x^2 - 2x + 1$,are you really sure about doing the division.
that is why i didn't do it

Originally Posted by Stroodle
Hi. The way I did it is:

$\displaystyle p(x)=ax^{2008}+bx^{2007}-1$

As previously stated, if $\displaystyle (x-1)$ is a factor, then $\displaystyle p(1)=0$

$\displaystyle \therefore a+b=1$

Now if you divide $\displaystyle P(x)$ by $\displaystyle (x-1)$ you get:

$\displaystyle g(x)=ax^{2007}+(b-a)x^{2006}+(a-b)x^{2005}+(b-a)x^{2004}+(a-b)^{2003}.......(b-a)-1$

Now $\displaystyle g(1)=0$ for $\displaystyle (x-1)$ to be a second factor.

Which gives $\displaystyle b-1=0$

$\displaystyle \therefore a=0$ and $\displaystyle b=1$
i think $\displaystyle g(x)$ should be
$\displaystyle g(x)=ax^{2007}+(b+a)x^{2006}+(a+b)x^{2005}+(b+a)x^ {2004}+(a+b)x^{2003}+ \cdot \cdot \cdot + (b+a)$

and

$\displaystyle g(1) = a + (b+a) + (a+b) + (b+a) + (a+b) + \cdot \cdot \cdot + (b+a) = 0$

or

$\displaystyle 2008a + 2007b = 0$

now, we have two equations in $\displaystyle a$ and $\displaystyle b$, solve it

13. im lost.. from that i get

b = 2007
a = -2006

im getting more and more confused.. what is wrong with my solution btw? :P

14. Originally Posted by metlx

$\displaystyle x_{1,2} = \frac{-b \pm \sqrt{b^2 -(- 4(1 - b))}}{2(1 - b)} = \frac{-b \pm \sqrt{b^2 - 4b + 4}}{2 - 2b}$
this formula only work for 2 degrees polynomial

15. Originally Posted by metlx
Find a and b (from reals) so that

$\displaystyle p(x) = ax^{2008} + bx^{2007} - 1$
will be dividable by polynomial $\displaystyle (x - 1)^2$.

no idea how to begin.. can someone help?
HI

In other words , $\displaystyle (x-1)^2$ would be a factor of $\displaystyle p(x)$ so $\displaystyle p(1)=0$

$\displaystyle p(1)=a(1)^{2008}+b(1)^{2007}-1$

$\displaystyle a+b=1$ --- 1

$\displaystyle p'(x)=2008ax^{2007}+2007bx^{2006}$

$\displaystyle p'(1)=0$

$\displaystyle 2008a+2007b=0$ --- 2

Solving the system would give you $\displaystyle a=-2007$ and $\displaystyle b=2008$