What is the irrational cube root ³√32 rewritten as the product of an integer and another irrational square root and simplest form
The irrational cube root can be rewritten as the product of an integer and another irrational square root in simplest form as follows:
³√32 = 2 * √2
Therefore, the cube root of 32 can be written as the product of an integer (2) and another irrational square root (√2) in simplest form.
To rewrite the irrational cube root ³√32 in the simplest form as the product of an integer and another irrational square root, we can start by finding the prime factorization of 32.
The prime factorization of 32 is 2 * 2 * 2 * 2 * 2, which can also be written as 2^5.
Now, let's rewrite ³√32 using this prime factorization:
³√32 = ³√(2^5)
Since the cube root operation distributes over multiplication, we can rewrite this as:
= ³√2^3 * ³√2^2
Taking the cube root of 2^3 gives us ²√2, and the cube root of 2^2 gives us ²√2. Therefore, we can write:
³√32 = ²√2 * ²√2^2
Simplifying further, we get:
³√32 = ²√2 * ²√4
Since 4 is a perfect square, we can simplify ²√4 as 2:
³√32 = ²√2 * 2
Therefore, the irrational cube root ³√32 can be rewritten as the product of an integer and another irrational square root in the simplest form as ²√2 * 2.
To rewrite the irrational cube root ³√32 as the product of an integer and another irrational square root in simplest form, we need to break down 32 into its prime factors.
First, we observe that 32 can be written as 2 × 2 × 2 × 2 × 2.
Now, let's rewrite 32 as the product of an integer and another irrational square root:
³√32 = ³√(2 × 2 × 2 × 2 × 2)
Since there are three groups of 2s in 32, we can take one group of 2s outside the cube root:
³√32 = ²√(2 × 2 × 2) × ³√2
Simplifying further:
³√32 = ²√(2³) × ³√2
By simplifying inside the square root, we have:
³√32 = 2² × ³√2
This can be rewritten as the product of an integer and an irrational square root in simplest form:
³√32 = 4√2