write the irrational square root √48 as the product of another irrational product and integer

√48 can be written as the product of the irrational square root of 16 and the integer 3.

√48 = √(16 * 3) = √16 * √3 = 4√3

To write the square root of 48 (√48) as the product of another irrational product and an integer, we need to find a perfect square factor of 48.

The prime factorization of 48 is:
48 = 2^4 * 3

We can see that 16 is a perfect square factor of 48, as 16 = 2^4. So we can divide 48 by 16 to get:
48/16 = 3

Therefore, we can write √48 as the product of the square root of 16 (√16) and the square root of 3 (√3) multiplied by 3:
√48 = √16 * √3 * 3

Simplifying further, we have:
√48 = 4√3

So √48 can be expressed as the product of another irrational product (√3) and an integer (4).

To write the irrational square root of √48 as the product of an irrational number and an integer, we first need to simplify the square root of 48.

We can start by factoring 48 into its prime factors: 48 = 2^4 * 3.

Now, let's look for perfect squares among these factors. We see that 4 is a perfect square, but 3 is not. So, we can simplify the square root of 48 by taking out the square root of 4:

√48 = √(4 * 2^2 * 3) = 2 * √(2^2 * 3).

Now, we can rewrite this expression as the product of an irrational number (√3) and an integer (2):

√48 = 2 * √(2^2 * 3) = 2√3.

Therefore, the irrational square root of √48 can be written as the product of the irrational square root of 3 and the integer 2:

√48 = 2√3.

what is the irrationl squre root of √28 simplified

To simplify the square root of √28, we need to find perfect square factors of 28.

The prime factorization of 28 is:
28 = 2^2 * 7

Since 4 is a perfect square factor of 28, we can simplify the square root of 28 by taking out the square root of 4:

√28 = √(4 * 7) = 2√7

Therefore, the irrational square root of √28 simplifies to 2√7.

rewrite the irrational cube root 3*√48 as a product of an integer and another irrational cube root

To rewrite the expression 3 * √48 as a product of an integer and another irrational cube root, we need to find a perfect cube factor of 48.

The prime factorization of 48 is:
48 = 2^4 * 3

We can see that 8 is a perfect cube factor of 48, as 8 = 2^3. So we can divide 48 by 8 to get:
48 / 8 = 6

Therefore, we can write 3 * √48 as the product of the cube root of 8 (∛8) and the square root of 6 (√6) multiplied by 3:
3 * √48 = 3 * (∛8 * √6)

Simplifying further, we have:
3 * √48 = 3 * ∛8 * √6

So, 3 * √48 can be expressed as the product of an irrational cube root (∛8) and an integer:
3 * √48 = 3∛8 * √6

rewrite the irrational cube root 3*√48 as a product of an integer and another irrational cube root

4 3*√12
-2 3*√6
2 3*√6
4 3*√3

To rewrite the expression 3 * √48 as a product of an integer and another irrational cube root, we need to find a perfect cube factor of 48.

The prime factorization of 48 is:
48 = 2^4 * 3

We can see that 27 is a perfect cube factor of 48, as 27 = 3^3. So we can divide 48 by 27 to get:
48 / 27 = 1 remainder 21

Therefore, we can write 3 * √48 as the product of the cube root of 27 (∛27) and the square root of 21 (√21) multiplied by the integer coefficient 1:
3 * √48 = 3 * (∛27 * √21)

Simplifying further, we have:
3 * √48 = 3 * ∛27 * √21

So, 3 * √48 can be expressed as the product of an integer (3) and another irrational cube root:
3 * √48 = 3∛27 * √21

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