For the polynomial $x^4+ax^3+bx^2+c=0$, with zeros $\alpha,\beta,\gamma,\text{ and }\sigma$, it is given that
(i) sum of all roots is $2$,
(ii) the product of all roots is $1$, and
(iii) the individually squared sum of roots is $0$.
Find the values of $a$, $b$, and $c$.
Can anyone explain how we're supposed to solve this, I've been stuck on it for an hour and I am relatively new to this topic
Let's start by factoring the polynomial into its roots. We can do this because of the Factor Theorem. $$x^4+ax^3+bx^2+c=(x-\alpha)(x-\beta)(x-\gamma)(x-\sigma)$$and by expanding this out we get $$=x^4-(\alpha+\beta+\gamma+\sigma)x^3+(\alpha\beta+\alpha\gamma+\alpha\sigma+\beta\gamma+\beta\sigma+\gamma\sigma)x^2-(\alpha\beta\gamma +\alpha\beta\sigma+\alpha\gamma\sigma+\beta\gamma\sigma)x+\alpha\beta\gamma\sigma$$(you can check this by doing the expansion manually and grouping like terms, but it's easier using something called Vieta's formula which is just a generalisation of this idea.) So now we can assign these coefficients of our original polynomial with the coefficients of the new one. We now get $$\begin{align*} -(\alpha+\beta+\gamma+\sigma)&=a\\ \alpha\beta+\alpha\gamma+\alpha\sigma+\beta\gamma+\beta\sigma+\gamma\sigma&=b\\ -(\alpha\beta\gamma +\alpha\beta\sigma+\alpha\gamma\sigma+\beta\gamma\sigma)&=0\\ \alpha\beta\gamma\sigma&=c \end{align*}$$ Then since you know some things about the sum, product and sum of individually squared roots, you can fairly easily use their relationships with coefficients $a,b,c$ to work out what those coefficients are.