MHB Finding $q$ in a Polynomial with Negative Integer Roots

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The discussion revolves around finding the value of $q$ in the polynomial $P(x)=x^4+mx^3+nx^2+px+q$, given that all roots are negative integers and the sum of the coefficients $m+n+p+q=2009$. Participants express appreciation for each other's solutions and engage in light conversation about sharing more math problems. The focus remains on the mathematical challenge presented by the polynomial and its constraints. The conversation highlights collaboration and problem-solving within the mathematical community.
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If $P(x)=x^4+mx^3+nx^2+px+q$ is a polynomial whose roots are all negative integers, and given that $m+n+p+q=2009$, find $q$.
 
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anemone said:
If $P(x)=x^4+mx^3+nx^2+px+q$ is a polynomial whose roots are all negative integers, and given that $m+n+p+q=2009$, find $q$.

Hello.

For my system:http://mathhelpboards.com/number-theory-27/polynomials-roots-10020.html

Let \ x_1, \ x_2, \ x_3, \ x_4: \ roots \ of \ P(x) \rightarrow{}1+m+n+p+q=2010

2010= Term independent from the polynomial, from roots: x_1-1, \ x_2-1, \ x_3-1, \ x_4-1

2010=2*3*5*67 \rightarrow{} x_1=-1, \ x_2=-2, \ x_3=-4, \ x_4=-66

P(x)=x^4+73x^3+476x^2+932x+528

Therefore:

q=528

Regards.
 
mente oscura said:

Hello.

For my system:http://mathhelpboards.com/number-theory-27/polynomials-roots-10020.html

Let \ x_1, \ x_2, \ x_3, \ x_4: \ roots \ of \ P(x) \rightarrow{}1+m+n+p+q=2010

2010= Term independent from the polynomial, from roots: x_1-1, \ x_2-1, \ x_3-1, \ x_4-1

2010=2*3*5*67 \rightarrow{} x_1=-1, \ x_2=-2, \ x_3=-4, \ x_4=-66

P(x)=x^4+73x^3+476x^2+932x+528

Therefore:

q=528

Regards.
Solution above is more structured
here is mine
if -a,-b,-c,-d are roots ( a, b, c,d all > 0) then expanding (x+a)(x+b)(x+c)(x+d) shen a+b + c + d = m
ab + bc + ac + ad + bd + cd = n
abc+ bcd+ acd + abd = p
abcd = q
now (1+a)(1+b)(1+c)(1+d) = 1 + ( a+b + c + d) + (ab + bc + ac + ad + bd + cd) + (abc+ bcd+ acd + abd) + abcd = 1+ m + n+ p + q = 2010 = 2 * 3 * 5 * 67
because of symetry considerions we can take a < b < c< d hence a = 1, b = 2, c= 4, d = 66
so q= abcd = 1 * 2 * 4 * 66 = 528
 
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Well done to both of you and thanks for participating! :cool:

Do you guys have also some intriguing math problems that you could share with us here? Hehehe...(Wink)
 
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