Polynomial question, am I correct?

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The discussion confirms that z - 1 is a factor of the polynomial p(z) = z^5 - 5z^4 + 18z^3 - 50z^2 + 81z - 45, as p(1) equals 0. The method for demonstrating this involves understanding that if a polynomial evaluates to zero at a certain point, then the corresponding linear factor is indeed a factor of the polynomial. The explanation highlights the relationship between polynomial division and the remainder theorem, clarifying why p(1) = 0 implies z - 1 is a factor. Overall, the reasoning presented is deemed correct and effectively illustrates the concept.
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[SOLVED] Polynomial question, am I correct?

Homework Statement


p(z) = z^{5}-5z^{4}+18z^{3}-50z^{2}+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^{5}-5z^{4}+18z^{3}-50z^{2}+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.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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