I need some help with this question: let $p,q,r,s$ be the roots of $3x^4 - x + 12 = 0$, find $pqr + pqs + prs + qrs$

Question

I need some help with this question: let $p,q,r,s$ be the roots of $3x^4 - x + 12 = 0$, find $pqr + pqs + prs + qrs$.

I know a few of Vieta's formulas but I don't know about the sum of the products of the roots for quartic equations or for any polynomial. I tried searching for it but couldn't find anything.

Answer 1

$$3x^4 - x + 12 = 3(x^4 - \frac{x}{3} + 4)$$Then, we know that $(x-p)(x-q)(x-r)(x-s) = x^4 - \frac{x}{3} + 4$. From here, you should be able to find the value of $pqr + pqs + prs + qrs$ without knowing the values of the roots themselves.

Answer 2

$$(x-p)(x-q)(x-r)(x-s)=0$$ $$x^4 -(p+q+r+s)x^3+(pq+pr+ps+qr+qs+rs)x^2-(pqr+pqs+prs+qrs)x+pqrs=0$$ Your equation is $3x^4-x+12=0$, divide by $3$ on both sides resulting in

$$x^4 - \frac{1}{3}x+4=0$$ Matching the coefficient for $x$, we'll get $$pqr+pqs+prs+qrs= \frac{1}{3}$$

Answer 3

Vieta's formulae tells us that

$${\displaystyle \sum _{1\leq i_{1}<i_{2}<\cdots <i_{k}\leq n}\left(\prod _{j=1}^{k}r_{i_{j}}\right)=(-1)^{k}{\frac {a_{n-k}}{a_{n}}}}$$

Since the polynomial in question is a quartic, we have that $n=4$. Thus, we have $${\displaystyle \sum _{1\leq i_{1}<i_{2}<i_{3} <i_{4}\leq 4}\left(\prod _{j=1}^{k}r_{i_{j}}\right)=\sum_{i=1}^4\frac{r_1r_2r_3r_4}{r_i} = \frac{a_4}{a_1} }$$

Looking at the polynomial, we see that $a_4 = 1$ and $a_1 = 3$, and thus

$$pqr+qrs+rsp + spq = \frac{a_4}{a_1} = \frac13$$