Factor the trigonometric expression completely.
$2 \cot^{2} W + 13 \cot W + 21$

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Answer

Final Answer: $(2 \cot (W) + 7) (\cot (W) + 3)$

Steps

Step 1 ：Given expression is $2 \cot^{2} (W) + 13 \cot (W) + 21$

Step 2 ：We need to find two numbers such that their product is $2 * 21 = 42$ and their sum is 13

Step 3 ：The numbers 6 and 7 satisfy these conditions because $6 * 7 = 42$ and $6 + 7 = 13$

Step 4 ：So, we can write the middle term as the sum of $6 \cot (W)$ and $7 \cot (W)$

Step 5 ： $2 \cot^{2} (W) + 13 \cot (W) + 21 = 2 \cot^{2} (W) + 6 \cot (W) + 7 \cot (W) + 21$

Step 6 ：Now, we can factor by grouping: $2 \cot (W) [\cot (W) + 3] + 7 [\cot (W) + 3]$

Step 7 ：Now, take out the common factor $[\cot (W) + 3]$

Step 8 ：So, the factorized form of the given expression is $(2 \cot (W) + 7) (\cot (W) + 3)$

Step 9 ：Final Answer: $(2 \cot (W) + 7) (\cot (W) + 3)$