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

[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)?

 

[neutrino]

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

 

[camilus]

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

 

[neutrino]

So what's your answer?

 

[camilus]

R(P+Q)=15

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

 

[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.

 

[neutrino]

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

 

[camilus]

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.

 

[neutrino]

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.

 

[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 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.

 

[uart]

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