How do i factorize this?
$\displaystyle
r^4-4r^3+7r^2-6r+2=0
$
It is $\displaystyle (r-1)^2(r^2-2r+2)=0$.
You can see that $\displaystyle r=1$ is a solution so you get $\displaystyle (r-1)(r^3-3r^2+4r-2)=0$ and you can realize that $\displaystyle r=1$ is a root of $\displaystyle r^3-3r^2+4r-2=0$ too, so you get $\displaystyle (r-1)^2(r^2-2r+2)=0$. And $\displaystyle r^2-2r+2=0$ hasn't got any real roots (because $\displaystyle D=2^2-4*2<0$) so you can't factorize that.
Hello, MilK!
I factored this at your other post,How do i factorize this?
$\displaystyle r^4-4r^3+7r^2-6r+2\:=\:0$
. . but here's how I did it.
Knowing the Rational Roots Theorem,
. . I knew that the only rational roots are: .$\displaystyle \pm1,\:\pm2$
I tried $\displaystyle r = 1:\;\;1^4 - 4(1^3) + 7(1^2) - 6(1) + 2 \;=\;0$
From the Remainder Theorem, I knew that $\displaystyle (r-1)$ is a factor.
Using long division: .$\displaystyle (r - 1)(r^3-3r^2+4r-2)$
I saw that $\displaystyle r = 1$ is a root of the cubic.
. . Hence, $\displaystyle (r-1)$ is a factor.
And we have: .$\displaystyle (r-1)(r-1)(r^2 -2r + 2)$
The quadratic does not factor; use the Quadratic Formula.
Have you heard of the rational root theorem?
-Dan