# Simple polynomial problem

1. Sep 10, 2007

### camilus

if the polynomial $$x^3+3x^2+9x+3$$ is a factor of $$x^4+4x^3+6Px^2+4Qx + R$$, what is $$R(P+Q)$$?

2. Sep 10, 2007

### neutrino

camilus, you need show your attempt before we can help.

3. Sep 10, 2007

### camilus

I already got the answer, I'm just trying to confirm it, see if I got the same answer other members get.

4. Sep 10, 2007

### neutrino

So what's your answer?

5. Sep 10, 2007

### camilus

R(P+Q)=15

Have you tried the problem yet? What did you get?

6. Sep 10, 2007

### neutrino

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.

7. Sep 10, 2007

### neutrino

Sorry, it was I who made a numerical error. I do get 15.

8. Sep 10, 2007

### camilus

Did you divide the first polynomial into the second?

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

9. Sep 10, 2007

### neutrino

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.

??? I don't understand.

10. Sep 10, 2007

### camilus

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 cant 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. Sep 13, 2007

### uart

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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