1. ## Factorize

How do i factorize this?

$
r^4-4r^3+7r^2-6r+2=0
$

2. Originally Posted by MilK
How do i factorize this?

$
r^4-4r^3+7r^2-6r+2=0
$
It is $(r-1)^2(r^2-2r+2)=0$.
You can see that $r=1$ is a solution so you get $(r-1)(r^3-3r^2+4r-2)=0$ and you can realize that $r=1$ is a root of $r^3-3r^2+4r-2=0$ too, so you get $(r-1)^2(r^2-2r+2)=0$. And $r^2-2r+2=0$ hasn't got any real roots (because $D=2^2-4*2<0$) so you can't factorize that.

3. Originally Posted by james_bond
It is $(r-1)^2(r^2-2r+2)=0$.
You can see that $r=1$ is a solution so you get $(r-1)(r^3-3r^2+4r-2)=0$ and you can realize that $r=1$ is a root of $r^3-3r^2+4r-2=0$ too, so you get $(r-1)^2(r^2-2r+2)$. And $r^2-2r+2=0$ hasn't got any real roots (because $D=2^2-4*2<0$) so you can't factorize that.
Sorry i am not gd at realization, is there any other way?

4. Hello, MilK!

How do i factorize this?

$r^4-4r^3+7r^2-6r+2\:=\:0$
I factored this at your other post,
. . but here's how I did it.

Knowing the Rational Roots Theorem,
. . I knew that the only rational roots are: . $\pm1,\:\pm2$

I tried $r = 1:\;\;1^4 - 4(1^3) + 7(1^2) - 6(1) + 2 \;=\;0$

From the Remainder Theorem, I knew that $(r-1)$ is a factor.

Using long division: . $(r - 1)(r^3-3r^2+4r-2)$

I saw that $r = 1$ is a root of the cubic.
. . Hence, $(r-1)$ is a factor.

And we have: . $(r-1)(r-1)(r^2 -2r + 2)$