Factoring Polynomial z^4-4z^3+6z^2-4z-15 =0

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The polynomial z^4-4z^3+6z^2-4z-15 can be factored using the Ruffini rule, initially yielding the factorization (z+1)(-z^3-5z^2+11z-15). Further attempts to factor the cubic polynomial led to the realization that dividing by (z-3) simplifies the process. The final solutions to the equation are -1, 3, and 1±2i. The discussion emphasizes the importance of correctly applying Ruffini's rule and checking the leading terms during polynomial division. Understanding these steps is crucial for successfully factoring polynomials and finding their roots.
Fabio010
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z^4-4z^3+6z^2-4z-15 =0

How can i factor this polynomial in order to find the solutions??


I tried with the ruffini' rule.

and i reached the following equation [(z+1)(-z^3-5z^2+11z-15)] =0

now how can i factor (-z^3-5z^2+11z-15) ?

i tried it, but i can not solve it... :/
 
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Proceed with Ruffini. You'll find another root (because the problem is easy) and the remaining factor is quadratic, whose solutions you get with the formula.
 
the solutions are

-1; 3; 1+/-2i

i am going to try with ruffini again.
 
i cant. Even knowing the solutions, i can not proceed with ruffini's rule.
Maybe something is escaping me.
 
Redo the quotient (z^4-4z^3+6z^2-4z-15)/(z+1), since the leading term must be z^3, not -z^3.
 
ok it now makes sense.

now i factor it


(z^3-5z^2+11z-15)/(z-3)



...

but without the solution i would never be able to discover that i should divide (z^3-5z^2+11z-15) by (z-3)
 
Do Ruffini again: try with the divisors of -15 of both signs.
 
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