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Perform the indicated operations.

$$(k+2)\left(12 k^{3}-3 k^{2}+k+1\right)$$

$(k+2)\left(12 k^{3}-3 k^{2}+k+1\right)=12 k^{4}+21 k^{3}-5 k^{2}+2$

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for this problem will take the simpler polynomial to and then distributed across the larger one. So that would be okay. Plus two times on this large room here. Right? So that would be 12 k to the fourth. My eyes three k. Cute. This case squared. Chris cake. Um, plus 24 k. A cute. When a six case screwed, it was two k plus two. By adding the two together would get 12 cage the fourth plus 21. Okay, cute minus five cakes squared close three k opposed to

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