Question

how do i get from f(x) = (ax+b)/(cx+d) to f(x) = a/c + (b-ad/c)(cx+d) please?

Answer 1

so from\[\frac{ax+b}{cx+d}\]to\[\frac{a}{c}+(b-\frac{ad}{c})(cx+d)\]?

Answer 2

yes, sorry, my latex skills are pathetic!

Answer 3

lol

Answer 4

Ahh... this took me a bit to figure out what was going on, but now I see it. OK, this is just a rather abstract polynomial division. What you are REALLY going to do here is to DIVIDE the binomials, and then write the result in the form:

\[\frac{ dividend }{ divisor }=quotient+\frac{ remainder }{ divisor }\]
No different than if you had:\[\frac{ 25 }{ 6 }=4+\frac{ 1 }{ 6 }\]

It's just UG.LI.ER because you are working all with variable.

To do the division of \(\dfrac{ ax+b }{ cx+d }\), I'm assuming that you'll want to use synthetic division, so you'll need to have a leading coefficient of 1 in that den'r, so first divide through the num'r and den'r by c:
\[\large \dfrac{ ax+b }{ cx+d }\cdot \frac{ \frac{ 1 }{c } }{ \frac{ 1 }{ c } }=\dfrac{ \frac{ a }{ c } x+\frac{ b }{ c } }{ x+\frac{ d }{ c } }\]

Now use that form in your synthetic division. Remember, that den'r is your \((x-root)\)

I know it looks bad, lol... but trust me, do the synthetic division carefully and you'll see everything fall into place!! :)

Answer 5

Oh WAIT ... except, what you have there is

\(\Large \dfrac{ dividend }{ divisor }=quotient+remainder \cdot divisor\).....??

Are you sure the problem is stated correctly?

Answer 6

I think the problem should be, show that:

\[\Large \dfrac{ ax+b }{ cx+d }=\dfrac{a}{c}+\dfrac{b-\frac{ad}{c}}{cx+d}\]

...because, well, THAT's true. So either there is a typo in the problem, or a misinterpretation of the problem as stated, here.