Question

rf

Answer 1

HI!!

Answer 2

this looks like it could be hard, but maybe not too hard

Answer 3

so its (x-5) at the bottom

Answer 4

yes

Answer 5

and the asymptote is \(y=x^2-3x\) so when you divide the top by the bottom you should get \(x^2-3x\) with some remainder

Answer 6

lets call the numerator \(p(x)\) so that
\[\frac{p(x)}{x-5}=x^2-3x+\frac{R}{x-5}\] or more simply
\[p(x)=(x^2-3x)(x-5)+R\]

Answer 7

now we can replace \(x\) by \(-1\) and set the result equal to \(3\) and see if we can solve for \(R\)

Answer 8

you with me so far?

Answer 9

cause we are almost done

Answer 10

yes

Answer 11

ok if we replace \(x\) by \(-1\) in the denominator, we get \(-1-5=-6\)

Answer 12

that means the numerator has to be \(-18\) so that the fraction will be
\[-\frac{18}{-6}=3\]

Answer 13

\[p(x)=(x^2-3x)(x-5)+R\]
put \(x=-1\) set the result equal to \(-18\) and solve for \(R\)

Answer 14

in other words, solve
\[p(-1)=((-1)^2-3(-1))(-1-5)+R=18\] for \(R|)

Answer 15

ooops \[p(-1)=((-1)^2-3(-1))(-1-5)+R=-18\]

Answer 16

there may have been an easier way to do it, but i don't know it
you get \[4\times (-6)+R=-18\] or
\[-24+R=-18\]

Answer 17

so what is the rational function

Answer 18

we get \(R=6\) right?

Answer 19

yep

Answer 20

that means the numerator is \[p(x)=(x^2-3x)(x-5)+6\]

Answer 21

multiply that out, combine like terms etc, and you answer will be that polynomial over \(x-5\)

Answer 22

if you do it correctly you will get \[\frac{x^3-8 x^2+15 x+6}{x-5}\]