Question

adding rational expressions:
what do I do with the 3v that's hanging out?
(v-2)
----------------- + 3v
(3v^4-15v^3-18v^2)

I already factored the denominator
I got 3v^2(v-6)(v+1),but I don't know what to do next

Answer 1

\[\frac{3v(3v^4-15v^3-18v^2)}{(3v^4-15v^3-18v^2)}\]

Answer 2

i multiply first? i thought i had to factor first?

Answer 3

To add fractions, they need to have the same common denominator
Multiply 3v by (3v^2(v-6)(v+1))/(3v^2(v-6)(v+1)) (this is technically equal to 1, so is allowed as it does not change the expression)
You'll get:

(v-2) + 3v (3v^2(v-6)(v+1))
---------------------------
3v^2(v-6)(v+1)

or
(v-2) + 9v^3(v-6)(v+1)
----------------------
3v^2(v-6)(v+1)

(This is keeping the denominator expression factored like you did, but is not really necessary as it doesn't help simplify anything else.

Answer 4

First step after multiplying 3v by the common denominator is actually this, if you were curious. I skipped it

(v-2) 3v (3v^2(v-6)(v+1))
------------------+ -----------------
3v^2(v-6)(v+1) 3v^2(v-6)(v+1)

Answer 5

i got 9v^3+6v^2-15v when i multiplied 3v and (3v^2(v-6)(v+1

Answer 6

i did this wrong... i am looking at my multiplication

Answer 7

i'll redo it

Answer 8

Think of \[3v^{2}(v-6)(v+1)\]
as three independent terms multiplied by eachother.
You have \[3v^{2}\] \[(v-6)\] and \[(v+1)\]
Multiplying it by 3v is just sticking another term on there in the multiplication. making it \[(3v)3v^{2}(v-6)(v+1)\]
You can simplyfy this by combining it with \[3v^{2}\] making \[9v^{3}\]

Answer 9

what am i doing wrong here??
3v * 3v^2 = 9v^3

3v * (v-6) = 3v^2-18v

3v(v+1)= 9v^2+3v

i get 9v^3+3v^2-18v+3v^2+3v
then i combine like terms and i get
9v^3+6v^2-15v

Answer 10

bump

Answer 11

derp i just multiplied it wrong, got it!