1. ## Factorize

How do i factorize this?

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

2. Originally Posted by MilK
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.

3. Originally Posted by james_bond
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 $\displaystyle (r-1)^2(r^2-2r+2)$. 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.
Sorry i am not gd at realization, is there any other way?

4. Hello, MilK!

How do i factorize this?

$\displaystyle r^4-4r^3+7r^2-6r+2\:=\:0$
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: .$\displaystyle \pm1,\:\pm2$

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.

5. Originally Posted by MilK
Sorry i am not gd at realization, is there any other way?
Have you heard of the rational root theorem?

-Dan

6. okay i got it, thanks everyone