I have a question which says that if the roots of the equation $$x^4-12x^3+cx^2+dx+81=0$$ are positive, then what is the value of $c$ and $d$.
I am currently doing High school algebra and haven't had any calculus training and I believe that this question involves the knowledge of differential calculus. Can anybody please help me with this question by using algebraic methods only. Any kind of hint would suffice, I just want to know where can I start up.
Assuming there are 4 roots (not 2 or 0) and assuming all coeficients are integers.
If the roots are $a, b ,c, d$ then $x^4 - 12x^3 + Cx^2 + Dx + 81 = (x-a)(x-b)(x-c)(x-d)$
$=x^4 - (a+b+c+d)x^3 +(ab + ac + ad + bc + bd+cd)x^2 - (abc + abd+acd+bcd)x + abcd$
So
1)$a+b+c+d = 12$
2)$ab + ac + ad + bc + bd + cd = C$
3)$abc + abd +acd+bcd = -D$
4)$abcd = 81$
Combining 1) and 4) there are only so many possiblities if we asssume $a,b,c,d$ are all positive integers. $a,b,c,d$ must be $1$ or multiples of $3$ as $abcd = 3^4$. But $a+b+c + d = 12$ so only one is as large as $9$ (because $9+9+1+1 > 12$). But if one is $9$ then two must be $3$ and $9 + 3+3+1 > 12$. So at most they are $3$. But there must be four $3$s so $3+3+3+3 = 12$ and ($a=b=c=d=3$).
[If that was a little too talky... consider if $abcd = 81$ then $\{a,b,c,d\} = $ $\{81,1,1,1\}$ or $\{27,3,1,1\}$ or $\{9,9,1,1\}$ or $\{9,3,3,1\}$ or $\{3,3,3,3\}$. $\{3,3,3,3\}$ is the only option where $a+b+c+d = 12$.]
So
$ab + ac + ad + bc + bd + cd = 9+9+9+9+9+9 = 54 =C$ and $abc + abd +acd+bcd = =27+27+27+27=108=-D; d=-108$
[and $x^4 - 12x^3 + 54x^2 -104x + 81= x^4 - 4*3*x^3 + 6*3^2*x^2 - 4*3^3*x + 3^4= (x-3)^4$]