anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{3v(3v^4-15v^3-18v^2)}{(3v^4-15v^3-18v^2)}\]
anonymous
  • anonymous
i multiply first? i thought i had to factor first?
anonymous
  • anonymous
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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)
anonymous
  • anonymous
i got 9v^3+6v^2-15v when i multiplied 3v and (3v^2(v-6)(v+1
anonymous
  • anonymous
i did this wrong... i am looking at my multiplication
anonymous
  • anonymous
i'll redo it
anonymous
  • anonymous
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 \[9v^{3}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
bump
anonymous
  • anonymous
derp i just multiplied it wrong, got it!

Looking for something else?

Not the answer you are looking for? Search for more explanations.