<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] It is true that $x=0$ is not a root of $ax^2+bx+c$, since if so then $c$ would be zero, and the second polynomial $cy^2+by+a$ is linear with at most one root. Similarly, $y=0$ is not a root of $cy^2+by+a$.
Consider
$$cy^2+by+a = y^2 \left(c+\frac by + \frac a{y^2}\right)$$
If $x=0$ is not a root of $ax^2+bx+c$, i.e. $p\ne 0$, then $y=\dfrac16$ and $y=\dfrac1p$ are the two roots of $cy^2+by+a$. Then
$$pqr = p\cdot \frac 16\cdot \frac1p$$
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