Question

roots

Answer 1

\[z^6-5z^5+6z^4+4z^3-8z^2=0\]

Answer 2

soo, u wanna find root ?

Answer 3

or solve for z :D

Answer 4

I factored out z^2 from it

Answer 5

That is very nice.. :)

Answer 6

ohh its linear , boring :P

Answer 7

\[\begin{align*}z^6-5z^5+6z^4+4z^3-8z^2&=z^2\left(z^4-5z^3+6z^2+4z-8\right)\\\\
&=z^2\left(z^2(z-2)(z-3)+4(z-2)\right)
\end{align*}\]

Answer 8

I was just kidding.. :P

Answer 9

\[z^2(z^4-5z^3+6z^2+4z-8)\]
\[z^2(z^2(z^2-5z+6)+4(z-2))\] ?

Answer 10

is that the step you skipped?

Answer 11

hmm, but the book says \[z^2(z+1)(z-2)^3\]

Answer 12

and 6 roots are 0,0,-1,2,2,2 @SithsAndGiggles

Answer 13

\[\begin{align*}z^6-5z^5+6z^4+4z^3-8z^2&=z^2\left(z^4-5z^3+6z^2+4z-8\right)\\\\
&=z^2\left(z^2\left(z^2-5z+6z\right)+4z-8\right)\\\\
&=z^2\left(z^2(z-2)(z-3)+4(z-2)\right)\\\\
&=z^2(z-2)\left(z^2(z-3)+4\right)\\\\
&=z^2(z-2)\left(z^3-3z^2+4\right)\\\\
&=z^2(z-2)\left(z^3-4z^2+4z+z^2-4z+4\right)\\\\
&=z^2(z-2)\left(z\left(z^2-4z^2+4\right)+z^2-4z+4\right)\\\\
&=z^2(z-2)\left(z\left(z-2\right)^2+(z-2)^2\right)\\\\
&=z^2(z-2)^3(z+1)
\end{align*}\]