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What is the irrational cube root 40 rewritten as a product of an integer and another irrational cube root?
O 8√5
O 2√5
O-2√5
O 2√10
To rewrite the irrational cube root of 40 as a product of an integer and another irrational cube root, we need to find a perfect cube factor of 40.
The perfect cube factors of 40 are 1, 8, and 27.
Since 40 is not a perfect cube, the only factor we can use to rewrite it as a product of an integer and another irrational cube root is 8.
Therefore, the correct answer is O 8√5.
To rewrite the irrational cube root of 40 as a product of an integer and another irrational cube root, we need to find a perfect cube that is a factor of 40.
The prime factorization of 40 is 2^3 * 5.
Since 2^3 = 8, we can rewrite the cube root of 40 as the cube root of 8 * 5.
Simplifying, we get ∛(8*5) = ∛8 * ∛5.
∛8 is equal to 2, so we can rewrite this as 2∛5.
Therefore, the answer is O 2√5.
To rewrite the irrational cube root of 40 as a product of an integer and another irrational cube root, we need to factorize 40 into its prime factors.
The prime factorization of 40 is 2 × 2 × 2 × 5 = 2³ × 5.
Now, let's look at each option:
Option A: 8√5. The prime factorization of 8 is 2 × 2 × 2 = 2³. So, rewriting the cube root of 40 as 8√5 makes it 2³ x √5 = 2√5, which is not the correct answer.
Option B: 2√5. The prime factorization of 2 is already accounted for in the prime factorization of 40. So, we can rewrite the cube root of 40 as 2 x √5, which matches the given option. Therefore, option B is the correct answer.
Option C: -2√5. The prime factorization does not have a negative sign, so option C is not correct.
Option D: 2√10. The prime factorization of 10 is 2 × 5, so if we rewrite the cube root of 40 as 2√10, it becomes 2 x √(2 × 5) = 2 x √10. Therefore, option D is not the correct answer.
Therefore, the correct answer is option B: 2√5.