Is R(P+Q) for This Polynomial Problem 15?

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Discussion Overview

The discussion revolves around determining the value of R(P+Q) for the polynomial problem involving the factorization of the polynomial x^3+3x^2+9x+3 into x^4+4x^3+6Px^2+4Qx + R. The scope includes mathematical reasoning and problem-solving techniques related to polynomial division and coefficient comparison.

Discussion Character

  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant states that R(P+Q) equals 15 and seeks confirmation from others.
  • Another participant expresses disagreement with the initial answer and suggests posting a complete solution to identify potential numerical errors.
  • A participant admits to making a numerical error and agrees that R(P+Q) is indeed 15.
  • There is a discussion about using polynomial long division to set coefficients equal to zero and resolve for P, Q, and R.
  • One participant proposes letting the other factor be (x+a) and sets up equations relating the coefficients, ultimately arriving at R=3, Q=3, and P=2, which also leads to R(P+Q)=15.

Areas of Agreement / Disagreement

There is some disagreement regarding the initial calculation of R(P+Q), but a later reply confirms that R(P+Q) equals 15. However, the method of arriving at this conclusion is debated, with different approaches suggested.

Contextual Notes

The discussion includes various methods for solving the polynomial problem, such as polynomial long division and coefficient comparison, but does not resolve which method is superior or universally applicable.

camilus
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if the polynomial [tex]x^3+3x^2+9x+3[/tex] is a factor of [tex]x^4+4x^3+6Px^2+4Qx + R[/tex], what is [tex]R(P+Q)[/tex]?
 
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camilus, you need show your attempt before we can help.
 
I already got the answer, I'm just trying to confirm it, see if I got the same answer other members get.
 
So what's your answer?
 
R(P+Q)=15

Have you tried the problem yet? What did you get?
 
I don't get the same answer as you do. But if you can post your complete solution, it would be better, since that would definitely indicate if you (or I) made a numerical error somewhere, or made a mistake in an earlier step.
 
Sorry, it was I who made a numerical error. I do get 15.
 
Did you divide the first polynomial into the second?

then you could get the separate results and set them equal to zero, or you could multiply back and create equations.
 
camilus said:
Did you divide the first polynomial into the second?
I didn't do that to solve the problem. But if you want to know if I double-checked the answer, then yes, it works out.

then you could get the separate results and set them equal to zero, or you could multiply back and create equations.

? I don't understand.
 
  • #10
Thats what I mean. Using long division of polynomials, the coefficients of each can be set equal to zero and resolved. Other than that, you can't multiply back the the (x+1) to the first polynomial and set the coefficients equal to the the coefficients of the second polynomial, and like earlier, just solve for P, Q, and R.
 
  • #11
camilus said:
Thats what I mean. Using long division of polynomials, the coefficients of each can be set equal to zero and resolved. Other than that, you can't multiply back the the (x+1) to the first polynomial and set the coefficients equal to the the coefficients of the second polynomial, and like earlier, just solve for P, Q, and R.

Hi camilus. My first instinct was to let the other factor be (x+a) and then just set up a few equations relating the coefficients.

That is,

r = 3a
4q = 9a+3
6p = 3a + 9
4 = a + 3

Since the last equation is trivial to solve (a=1) then solutions for r, p and q follow immediately.

So I got r = 3, q = 3 and p = 2, giving r(p+q) = 15
 

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