What is the irrational cube root −24−−−−√3 rewritten as a product of an integer and another irrational cube root?(1 point) Responses −23–√3 negative 2 root index 3 start root 3 end root 26–√3 2 root index 3 start root 6 end root 23–√3 2 root index 3 start root 3 end root −24–√3
To rewrite the irrational cube root −24−−−−√3 as a product of an integer and another irrational cube root, we need to simplify the expression.
First, let's simplify the cube root of −24:
- The cube root of 24 is 2 because 2 × 2 × 2 = 8, which is less than 24.
- Since we need the negative, the cube root of −24 is −2.
Now, let's simplify the cube root of √3:
- The cube root of 3 is 1 because 1 × 1 × 1 = 1, which is less than 3.
- Since we have a square root of 3, the cube root of √3 will be the cube root of 3 times the square root of 3.
So, the cube root of √3 can be written as 1 times √3.
Finally, we can rewrite the original expression as a product of an integer and another irrational cube root:
−24−−−−√3 = −2√3
Hence, the correct option is: −2√3
The correct response is: −23−√3
To rewrite the irrational cube root of −24−−−−√3 as a product of an integer and another irrational cube root, we can factor out the largest perfect cube that is a factor of −24.
The largest perfect cube factor of −24 is −8 (because −8 x −8 x −8 = −24).
This gives us:
−24−−−−√3 = −8√3 * (3−−−−√3)
So, the irrational cube root −24−−−−√3 can be rewritten as a product of an integer (-8) and another irrational cube root (3−−−−√3).