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?
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?
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
that is why i didn't do it
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
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 $\displaystyle b=2008$