<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] Considering roots of $ax^2+bx+c$ as $\frac{-b\pm \delta}{2a}$ the roots of $cx^2+bx+a$ will be $\frac{-b\pm \delta}{2c}$. So, the product of all the four roots are $$6PQR=\frac{(-b+ \delta)(-b-\delta)}{2a}\times\frac{(-b+ \delta)(-b-\delta)}{2c}=\frac{4ac}{4ac}=1$$ Thus, we have $$PQR=\frac16$$
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