what is the irrational cube root ^3√32 rewritten as a product of an integer and another irrational cube root?(1 point)
A.2 ^3√4
B.4 ^3√8
C. 8 ^3√4
D.4 ^3√2
The cube root of 32 is the number that, when multiplied by itself three times, equals 32. 32 can be expressed as the product of 2 and 16, with 2 being a perfect cube.
So, we can rewrite the cube root of 32 as:
^3√32 = ^3√(2 * 16)
By using the property of cube roots, ^3√(a * b) = ^3√a * ^3√b, we can rewrite it as:
^3√32 = ^3√2 * ^3√16
So the answer is A.2 ^3√4
To rewrite the irrational cube root ^3√32 as a product of an integer and another irrational cube root, we need to find a perfect cube that is a factor of 32.
The perfect cube factors of 32 are 2 and 8.
Therefore, ^3√32 can be rewritten as 2 × ^3√4, since 4 = 2^2.
So the correct answer is A. 2 ^3√4.
To rewrite the cube root of 32 as a product of an integer and another irrational cube root, we need to find a perfect cube that can be extracted from 32.
First, let's factorize the number 32:
32 = 2 * 2 * 2 * 2 * 2
Since we are looking for a perfect cube, we need to group the factors in sets of three:
32 = (2 * 2 * 2) * (2 * 2)
We can rewrite this as:
32 = 8 * 4
Now, we can express 8 as a perfect cube multiplied by another cube root:
8 = (2 * 2 * 2) = 2^3
Therefore, we have:
∛32 = ∛(8 * 4) = ∛8 * ∛4 = 2∛4
So, the irrational cube root ∛32 can be rewritten as the product of the integer 2 and the irrational cube root ∛4.
Therefore, the correct answer is:
A. 2∛4