1. ## factorization

Given $p(x)=x^3+2x^2-x-2$

If $q(y)=y^6+2y^5-4y^4-6y^3+4y^2+2y-1$ , find the real number k such that
$

q(y)=y^3[(y-\frac{1}{y})^3+2(y-\frac{1}{y})^2-(y-\frac{1}{y})+k]
$
, y is not 0 .

By using the substitution $x=y-\frac{1}{y}$ , show that the equation $q(y)=0$ can be transformed into the equation $p(x)=0$ .

THanks ..

2. Originally Posted by thereddevils
Given $p(x)=x^3+2x^2-x-2$

If $q(y)=y^6+2y^5-4y^4-6y^3+4y^2+2y-1$ , find the real number k such that
$

q(y)=y^3[(y-\frac{1}{y})^3+2(y-\frac{1}{y})^2-(y-\frac{1}{y})+k]
$
, y is not 0 .

By using the substitution $x=y-\frac{1}{y}$ , show that the equation $q(y)=0$ can be transformed into the equation $p(x)=0$ .

THanks ..
The constant term in $\left(y-\frac{1}{y}\right)^3+2\left(y-\frac{1}{y}\right)^2-\left(y-\frac{1}{y}\right)+k$ is $-4+k$ and this has to equal the coefficient of $y^3$ in $q(y).$ Hence $-4+k=-6\ \implies\ k=-2.$

3. Originally Posted by TheAbstractionist
The constant term in $\left(y-\frac{1}{y}\right)^3+2\left(y-\frac{1}{y}\right)^2-\left(y-\frac{1}{y}\right)+k$ is $-4+k$ and this has to equal the coefficient of $y^3$ in $q(y).$ Hence $-4+k=-6\ \implies\ k=-2.$

THanks .But i don understand how did u get the constant term ?

4. $\left(y-\frac{1}{y}\right)^3$ has no constant term (the terms are in $y^3,\,y,\,y^{-1}$ and $y^{-3}).$ Neither does $-\left(y-\frac{1}{y}\right)$ have any constant term. Hence the only constant term besides the $k$ comes from $2\left(y-\frac{1}{y}\right)^2$ and that is -4.

As a last resort, you can expand out $\left(y-\frac{1}{y}\right)^3+2\left(y-\frac{1}{y}\right)^2-\left(y-\frac{1}{y}\right)+k$ completely and see what happens.

5. hi
i have the same problem with 3 constants to find
Could you please show me how to find them with the same method

$f : x \rightarrow \frac{x^{2}+x-2}{x^{2}+x}$
find a,b and c such :
$f(x) = a + \frac{b}{x}+\frac{c}{x+1}$
$x$ is not 0 nor -1
thanks a lot.

6. Originally Posted by Raoh
hi
i have the same problem with 3 constants to find
Could you please show me how to find them with the same method

$f : x \rightarrow \frac{x^{2}+x-2}{x^{2}+x}$
find a,b and c such :
$f(x) = a + \frac{b}{x}+\frac{c}{x+1}$
$x$ is not 0 nor -1
thanks a lot.

$f:\rightarrow \frac{x^2+x-2}{x^2+x}$

so

$f(x) = a+\frac{b}{x} + \frac{c}{x+1} = \frac{x^2+x-2}{x^2+x}$

$a+\frac{b}{x} + \frac{c}{x+1} = \frac{x^2+x}{x^2+x} + \frac{-2}{x^2+x}$

$a+\frac{b}{x} + \frac{c}{x^2+x} = 1 + \frac{-2}{x^2+x}$

$\frac{-2}{x^2+x} =\frac{A}{x} + \frac{B}{x+1}$

$-2= A(x+1) + Bx$

$A=-2 , B=2$

$a+\frac{b}{x} + \frac{c}{x+1}=1+\frac{-2}{x} + \frac{2}{x+1}$

it is clear

7. thanks a lot
i liked how "TheAbstractionist" answered the first problem,is there any way we can do this one with the same method.
thanks again