'Equating the coefficients' question

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SUMMARY

The discussion centers on expanding the expression (x-3)(px^2+qx+r) to derive a cubic polynomial. The correct expansion results in px^3 + (q-3p)x^2 + (r-3q)x - 3r. Participants clarify the process of collecting like terms after expansion, emphasizing the importance of recognizing the structure of the cubic polynomial. The final expression demonstrates how coefficients are adjusted during the factorization process.

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  • Understanding polynomial expansion techniques
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  • Knowledge of collecting like terms in algebra
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Students studying algebra, particularly those focusing on polynomial functions and their expansions, as well as educators seeking to clarify polynomial manipulation techniques.

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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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