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

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.

• Have you tried going on the wikipedia page? Commented Dec 18, 2023 at 3:59
• Then you may try to expand $3(x-p)(x-q)(x-r)(x-s)$. Commented Dec 18, 2023 at 4:02
• Take a look here. Commented Dec 18, 2023 at 4:05

$$3x^4 - x + 12 = 3(x^4 - \frac{x}{3} + 4)$$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.

• I think you mean $x^4$, not $x^2$. Commented Dec 18, 2023 at 4:17
• Oops, Thank you! Commented Dec 18, 2023 at 4:18

Vieta's formulae tells us that

$${\displaystyle \sum _{1\leq i_{1}

Since the polynomial in question is a quartic, we have that $$n=4$$. Thus, we have $${\displaystyle \sum _{1\leq i_{1}

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

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

$$(x-p)(x-q)(x-r)(x-s)=0$$ $$x^4 -(p+q+r+s)x^3+(pq+pr+ps+qr+qs+rs)x^2-(pqr+pqs+prs+qrs)x+pqrs=0$$ 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 $$pqr+pqs+prs+qrs= \frac{1}{3}$$