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$x^2-4x-12$ 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.

Wang Kah Lun
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saoceallaigh
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1 Answers1

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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 $s$.

Wang Kah Lun
  • 10,240
  • Thanks https://math.stackexchange.com/users/165867/alan-wang for your much simpler and more direct solution. However that does not address my problems with the more painful long division method as presented by the author of the problem?? – saoceallaigh Jun 10 '19 at 18:55
  • @saoceallaigh Updated. Inform me if you still have problem using long division. – Wang Kah Lun Jun 11 '19 at 00:38
  • https://math.stackexchange.com/users/165867/alan-wang No All good now. Thanks for all your help – saoceallaigh Jun 11 '19 at 09:29
  • @saoceallaigh Welcome. And if you accept my answer, please press the tick on the left hand side of my answer. Thank you. – Wang Kah Lun Jun 11 '19 at 17:18