'Equating the coefficients' question

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The discussion revolves around expanding the expression (x-3)(px^2+qx+r) to obtain a cubic polynomial. The initial expansion leads to the terms px^3, qx^2, and rx, which are then combined with the results from distributing -3 across the other terms. The confusion arises in understanding how to combine and factor the resulting terms into the form px^3+(q-3p)x^2+(r-3q)x-3r. Clarification is provided that recognizing the structure of the cubic polynomial is essential for proper factorization. The participant acknowledges the need to improve their ability to identify these patterns in polynomial expressions.
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Homework Statement


Expand the brackets to get a cubic containing the unknowns. This is an example in the textbook but I don't see how they've expanded the brackets to get their answer:

(x-3)(px^2+qx+r)=px^3+(q-3p)x^2+(r-3q)x-3r

Homework Equations


(x-3)(px^2+qx+r)

The Attempt at a Solution



I would have done it how you expand long brackets:

(x-3)(px^2)+(x-3)(qx)+(x-3)(r)=px^3-3px^2...
 
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Yes, and the next step is to collect all the terms involving x^2 together, and all the terms involving x together...
 
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pasmith said:
Yes, and the next step is to collect all the terms involving x^2 together, and all the terms involving x together...

What I don't get is how they factorized what you get from expanding: px^3+qx^2+rx-3px^2-3qx-3r. How do you get px^3+(q-3p)x^2+(r-3q)x-3r from that?
 
Ah, I see if you do (q-3p)x^2 and multiply it out and the same with the other factorized terms, you get the terms you get when you'd first expand it out. I hadn't noticed that px^3+(q-3p)x^2... etc. was a cubic. I need to work on noticing things, I've found. :rolleyes:
 
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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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