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neverforgetvivistee
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
\[\frac{3v(3v^4-15v^3-18v^2)}{(3v^4-15v^3-18v^2)}\]
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.
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
derp i just multiplied it wrong, got it!