Finding $q$ in a Polynomial with Negative Integer Roots

[anemone]

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 $q$.

 

[mente oscura]

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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 $q$.

Hello.

Spoiler

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.

 

[kaliprasad]

​

Hello.

Spoiler

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

Spoiler

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

 

[anemone]

Well done to both of you and thanks for participating!

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