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Factor completely.
2k  k3  3k2
k(k + 1)(k  2)
k(k  1)(k  2)
k(k + 1)(k + 2)
 one year ago
 one year ago
Factor completely. 2k  k3  3k2 k(k + 1)(k  2) k(k  1)(k  2) k(k + 1)(k + 2)
 one year ago
 one year ago

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CallistoBest ResponseYou've already chosen the best response.1
What is the common factor of the three terms there?
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Hmm.. One more.. Can you see the negative signs?
 one year ago

Shannon20150Best ResponseYou've already chosen the best response.0
is it the 2nd one
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Yup! So, take out the common factor and group the rest of the terms. You'll get this: \[2k  k^3  3k^2\]\[=k(2+ k^2 + 3k)\]Rearrange the terms a bit: \[ = k (k^2+3k+2)\] From this, we can see that the 2nd one is nowhere near to the answer.
 one year ago
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