Simplifying fractions

[Pinkjess182]

I'm a ninth grader working on an algebra 1 packet. The directions of the section I'm stuck on say:
write each expression in simplest form. The problem is this:

28c²d-84c²
35cd-105c

I factored the numbers and came up with this:

28c(cd-3c)
35c(d-3)
I'm stuck because I don't know what to do next. Thanks for the help!

 

[soroban]

Hello, Pinkjess182!

You didn't factor completely.

Write the expression in simplest form: .\(\displaystyle \dfrac{28c^2d-84c^2}{35cd-105c}\)

Factor completely: .\(\displaystyle \dfrac{28c^2(d-3)}{35c(d-3)} \;=\;\dfrac{28c^2(\rlap{/////}d-3)}{35c(\rlap{/////}d-3)} \;=\;\dfrac{28c^2}{35c} \;=\;\dfrac{4c}{5}\)

Edit: corrected my errors.
. . . .Thanks for the heads-up, Yogi.

 

[Yogi]

Hello, Pinkjess182!

You didn't factor completely.


Factor completely: .\(\displaystyle \dfrac{28c^2(d-3c)}{35c(d-3c)} \;=\;\dfrac{28c^2(\rlap{//////}d-3c)}{35c(\rlap{//////}d-3c)} \;=\;\dfrac{28c^2}{35c} \;=\;\dfrac{4c}{5}\)

You have some extra c's in there.
\(\displaystyle \dfrac{28c^2(d-3)}{35c(d-3)} \;=\;\dfrac{28c^2(\rlap{//////}d-3)}{35c(\rlap{//////}d-3)} \;=\;\dfrac{28c^2}{35c} \;=\;\dfrac{4c}{5}\)