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

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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

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\[\frac{3v(3v^4-15v^3-18v^2)}{(3v^4-15v^3-18v^2)}\]
i multiply first? i thought i had to factor first?
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.

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

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