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Shannon20150
 2 years ago
Factor completely.
2k  k3  3k2
k(k + 1)(k  2)
k(k  1)(k  2)
k(k + 1)(k + 2)
Shannon20150
 2 years ago
Factor completely. 2k  k3  3k2 k(k + 1)(k  2) k(k  1)(k  2) k(k + 1)(k + 2)

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Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1What is the common factor of the three terms there?

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm.. One more.. Can you see the negative signs?

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1Yup! 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.
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