Polynomial question, am I correct?

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Homework Help Overview

The discussion revolves around a polynomial function p(z) = z^{5}-5z^{4}+18z^{3}-50z^{2}+81z-45 and the task of demonstrating that z - 1 is a factor of this polynomial.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between roots and factors of polynomials, specifically referencing the evaluation of p(1) to establish that z - 1 is a factor. There is a question regarding the correctness of the method used to demonstrate this.

Discussion Status

Some participants express confidence in the conclusion that z - 1 is a factor based on the evaluation of p(1), while others seek clarification on the method used to arrive at this conclusion. There is an acknowledgment of a fundamental principle regarding polynomial division and factors.

Contextual Notes

There are no explicit constraints mentioned, but the discussion implies a focus on verifying the method of proof within the context of polynomial factorization.

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[SOLVED] Polynomial question, am I correct?

Homework Statement


p(z) = z[tex]^{5}[/tex]-5z[tex]^{4}[/tex]+18z[tex]^{3}[/tex]-50z[tex]^{2}[/tex]+81z-45
Show that z - 1 is a factor of p(z)


Homework Equations



None

The Attempt at a Solution



Well since 1 is a root of p(z), then p(1) = 0
Then z - 1 is a factor?
 
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JFonseka said:

Homework Statement


p(z) = z[tex]^{5}[/tex]-5z[tex]^{4}[/tex]+18z[tex]^{3}[/tex]-50z[tex]^{2}[/tex]+81z-45
Show that z - 1 is a factor of p(z)


Homework Equations



None

The Attempt at a Solution



Well since 1 is a root of p(z), then p(1) = 0
Then z - 1 is a factor?

p(1) is equal to 0, than yes it is a factor.
 
I know it is a factor, but is my method correct for 'showing' that it is so?
 
That was PhY's answer! (less a stray "e"). Because p(1) is equal to 0, then yes it is a factor.

You are using a very simple fact about polynomials- If P(x) is any polynomial, dividing P(x) by (x-a) results in a quotient, Q(x), and a remainder, r, which, since x-a has degree 1, must be a number: P(x)= Q(x)(x-a)+ r. Now, let x= a: P(a)= Q(x)(a- a)+ r= r. If P(a)= 0 then r must be 0 and so (x- a) is a factor of P(x).
 
Thanks for clearing that up.
 

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