<s>[INST] The roots of $ax^2+bx+c$ are 6 and $P$. The roots of $cx^2+bx+a$ are $Q$ and $R$ what is the value of $P\times Q\times R$ Problem
The roots of $ax^2+bx+c$ are 6 and $P$. The roots of $cx^2+bx+a$ are $Q$ and $R$
And we are asked to find $P\times Q\times R$ by using the identities:
$P(x)=Q(x)\times D(x)+R(x)$
where $P(x)$ is the equations and $Q(x)$ is the quotient and $D(x)$ is the divisor and $R(x)$ is the remainder.
My thought process
so I had set $$ax^2+bx+c=Q(x)\times(x-6)\times(x-P)+0$$
and in the next equation:
$$cx^2+bx+a=Q(x)\times(x-P)(x-Q)+0$$
I really don't know where to go from here
 [/INST] $$ax^{ 2 }+bx+c=0\\ 6p=\frac { c }{ a } \\ cx^{ 2 }+bx+a=0\\ \\ qr=\frac { a }{ c } \\ 6pqr=1\Rightarrow \quad pqr=\frac { 1 }{ 6 } $$
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