I'm trying to solve a question that says that for the equation $z^4 + pz^3 + 54z^2 - 108z + 80 = 0$, three of the roots of the equation are $3 + i, a$ and $a^2$.
I know that for a quadratic equation $ax^2 + bx + c = 0$, the product of the roots is $\frac{c}a$, but I don't know how this would apply to the polynomial equation in this question.
Please help me, thank you!
On
For a cubic case if we let the roots be $a,b,c$,
$$ k\left(x-a\right)\left(x-b\right)\left(x-c\right)\\~\\ =k\left(x\left(x-b\right)\left(x-c\right)-a\left(x-b\right)\left(x-c\right)\right)\\~\\ =k\left(x^2\left(x-c\right)-xb\left(x-c\right)-ax\left(x-c\right)+ab\left(x-c\right)\right)\\~\\ =k\left(x^3-cx^2-bx^2+bcx-ax^2+acx+abx-abc\right)\\~\\ =k\left(x^3-ax^2-bx^2-cx^2+abx+bcx+acx-abc\right)\\~\\ =kx^3-k(a+b+c)x^2+k(ab+bc+ac)x-kabc\\~\\ $$
as you can see, the last term is the product of the roots times the coefficient of the first term. This applies generally to any polynomial where if $p\left(x\right)$ is a polynomial of degree $n$, the product of roots will be $$\frac{(-1)^na_0}{a_n}$$So your product of roots is $80$.
From Vieta's formula, for polynomial $p(x) = \sum_{i=1}^n a_i x^n$ and $r_i$ are the roots, we have $$\prod_{i=1}^nr_i = (-1)^n\frac{a_0}{a_n} $$
Hence the produce is $80$. Note that there is no need to compute the roots in order to obtain the product of the roots.