Eqyivalent expresion to 3√m^2n^5
1 answer:
The first step to solving this expression is to factor out the perfect cube
![\sqrt[3]{m^{2} n^{3} X n^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bm%5E%7B2%7D%20%20n%5E%7B3%7D%20X%20n%5E%7B2%7D%20%20%20%7D%20)
The root of a product is equal to the product of the roots of each factor. This will make the expression look like the following:
![\sqrt[3]{ n^{3} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20n%5E%7B3%7D%20%7D%20)
![\sqrt[3]{ m^{2} n^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20m%5E%7B2%7D%20n%5E%7B2%7D%20%20%7D%20)
Finally,, reduce the index of the radical and exponent with 3
n
This means that the correct answer to your question is n![\sqrt[3]{ m^{2} n^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20m%5E%7B2%7D%20n%5E%7B2%7D%20%7D%20)
.
Let me know if you have any further questions
:)
The root of a product is equal to the product of the roots of each factor. This will make the expression look like the following:
Finally,, reduce the index of the radical and exponent with 3
n
This means that the correct answer to your question is n
Let me know if you have any further questions
:)
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Sent a picture of the solution to the problem (s).
Given:
Polynomial is .
To find:
The sum of given polynomial and the square of the binomial (x-8) as a polynomial in standard form.
Solution:
The sum of given polynomial and the square of the binomial (x-8) is
On combining like terms, we get
Therefore, the sum of given polynomial and the square of the binomial (x-8) as a polynomial in standard form is .