<s>[INST] Finding unknowns in polynomial with two factors and remainder. $x^2-4x-12$ is a factor of $rx^3-sx^2+36$, find $r$ and $s$.
Long division gives $rx+(-s+4r)$ with remainder $12rx+4(-s+4r)x+36+12(-s+4r)$
Where I have difficulty with the logic  is that since $x^2-4x-12=0$ then the remainder is 0?
And from that we can say(how?) that $12r+4(-s+4r)=0$
And $36+12(-s+4r)=0$ also.
The rest is straightforward.
 [/INST] 
Factor Theorem
$x-a$ is a factor of $p(x)$ iff $p(a)=0$

Since $x^2-4x-12=(x-6)(x+2)$ is a factor of $p(x)=rx^3-sx^2+36$,
we know that $x-6$ and $x+2$ are factors of $p(x)$.
Thus $p(6)=0$ and $p(-2)=0$, that is $216r-36s+36=0$ and $-8r-4s+36=0$.
Solving the simultaneous linear equations give you $r=1,s=-7$.
To use long division, you need to divide $p(x)$ by $x-6$ and $x+2$.
By dividing $x-6$, you will get the remainder $216r-36s+36$.
By dividing $x+2$, you will get the remainder $-8r-4s+36$.
Since both are factors of $p(x)$, the remainder should be $0$.
Thus you get $216r-36s+36=0$ and $-8r-4s+36=0$.
If you straight away divide $p(x)$ by $x^2-4x-12$, the remainder you get is $12r+4(-s+4r)x+36+12(-s+4r)$.
Since the remainder must be zero polynomial,
you have $12r+4(-s+4r)=0$ and $36+12(-s+4r)=0$,
However, these two equations are multiple of one another so they cannot give you the solutions for $r$ and $s$.
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