Dividing single term algebraic fractions that won't factorise

[879282]

I've been struggling with a question for a while, which has come up several times in different forms on mock exams I've been doing in preparation for an exam next week. I can't find any info on how to do it online so I'm hoping someone can help.

An example question:
1) Divide
6j
9-8j

There's no solutions so I have no idea how to do this. I know how to do these questions by cancelling terms or by factorising and using partial fractions, but this question doesn't allow any of that.

Anyone that can show me how to do this, it is much appreciated

 

[Dr.Peterson]

An example question:
1) Divide
6j
9-8j

There's no solutions so I have no idea how to do this. I know how to do these questions by cancelling terms or by factorising and using partial fractions, but this question doesn't allow any of that.

Is this all that is said?

Could the context be such that "j" is the imaginary unit (also called i)? That is particularly common in engineering.

If j is just a variable, then all I can see to do is polynomial division, 6j divided by (-8j + 9). But it is not clear that that is what is intended.

 

[HallsofIvy]

I've been struggling with a question for a while, which has come up several times in different forms on mock exams I've been doing in preparation for an exam next week. I can't find any info on how to do it online so I'm hoping someone can help.

An example question:
1) Divide
6j
9-8j

There's no solutions so I have no idea how to do this. I know how to do these questions by cancelling terms or by factorising and using partial fractions, but this question doesn't allow any of that.

Anyone that can show me how to do this, it is much appreciated

If j here is the "imaginary unit" then multiply both numerator and denominator by the complex conjugate of the denominator 9+ 8j. That gives [math]\frac{6j}{9- 8j}\frac{9+ 8j}{9+ 8j}= \frac{54j+ 48}{81+ 64}= \frac{54}{150}j+ \frac{48}{150}= \frac{24}{75}+ \frac{9}{25}j [/math].

If j is simply a variable, then [math]\frac{6j}{9- 8j}= -\frac{3}{4}+ \frac{\frac{27}{4}}{9- 8j}[/math]. Continuing the division you could get an infinite sum of negative powers of j.

 

[lookagain]

This site does not use [math], [/math] for Latex.

Here is what you intended:

If j here is the "imaginary unit" then multiply both numerator and denominator by the complex conjugate of the denominator 9+ 8j. That gives \(\displaystyle \frac{6j}{9- 8j}\frac{9+ 8j}{9+ 8j}= \frac{54j+ 48}{81+ 64}= \frac{54}{150}j+ \frac{48}{150}= \frac{24}{75}+ \frac{9}{25}j
\)

If j is simply a variable, then \(\displaystyle \frac{6j}{9- 8j}= -\frac{3}{4}+ \frac{\frac{27}{4}}{9- 8j}. \ \ \ \) Continuing the division you could get an infinite sum of negative powers of j.