Answer
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Hint: In this particular question use the concept that rationalizing is a technique such that the $ {n^{th}} $ root polynomial is convert into a simple polynomial i.e. without $ {n^{th}} $ root or any other root by multiplying and divide by the polynomial such that the numerator root is cancel out.
Complete step-by-step answer:
Given polynomial:
$ \sqrt[5]{{{a^2}{b^3}{c^4}}} $
Now we have to find out the rationalizing factor of this polynomial.
Now as we know that rationalizing is a technique such that the $ {n^{th}} $ root polynomial is convert into a simple polynomial i.e. without $ {n^{th}} $ root or any other root by multiplying and divide by the polynomial such that the numerator root is cancel out.
For example consider a simple square root polynomial $ \sqrt x $ so to convert into a simple polynomial we have to multiply and divide by $ \sqrt x $ such that the numerator root is canceled out.
$ \Rightarrow \sqrt x = \sqrt x \times \dfrac{{\sqrt x }}{{\sqrt x }} = \dfrac{{\sqrt {{x^2}} }}{{\sqrt x }} = \dfrac{x}{{\sqrt x }} $
So here $ \sqrt x $ is the rationalizing factor.
Now given polynomial is $ \sqrt[5]{{{a^2}{b^3}{c^4}}} $
Now as we see that it is 5th root of $ {a^2}{b^3}{c^4} $ so we have to multiply and divide by 5th root of $ {a^3}{b^2}c $ such that its numerator root is canceled out.
\[ \Rightarrow \sqrt[5]{{{a^2}{b^3}{c^4}}} = \sqrt[5]{{{a^2}{b^3}{c^4}}} \times \dfrac{{\sqrt[5]{{{a^3}{b^2}c}}}}{{\sqrt[5]{{{a^3}{b^2}c}}}} = \dfrac{{\sqrt[5]{{{a^5}{b^5}{c^5}}}}}{{\sqrt[5]{{{a^3}{b^2}c}}}} = \dfrac{{abc}}{{\sqrt[5]{{{a^3}{b^2}c}}}}\]
Therefore, rationalizing factor of $ \sqrt[5]{{{a^2}{b^3}{c^4}}} $ is \[\sqrt[5]{{{a^3}{b^2}c}}\].
So this is the required answer.
Hence option (A) is the correct answer.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall that the rationalizing factor is a term by which when we multiply and divide the given polynomial its numerator becomes a simple polynomial, so first check the root of the given polynomial then check the power of the individual variables then multiply and divide by the same root of the polynomial and inside the root the power of the variable is such that the sum of the power of the same variables is equal to the root number such that variables can easily come out from the given root.
Complete step-by-step answer:
Given polynomial:
$ \sqrt[5]{{{a^2}{b^3}{c^4}}} $
Now we have to find out the rationalizing factor of this polynomial.
Now as we know that rationalizing is a technique such that the $ {n^{th}} $ root polynomial is convert into a simple polynomial i.e. without $ {n^{th}} $ root or any other root by multiplying and divide by the polynomial such that the numerator root is cancel out.
For example consider a simple square root polynomial $ \sqrt x $ so to convert into a simple polynomial we have to multiply and divide by $ \sqrt x $ such that the numerator root is canceled out.
$ \Rightarrow \sqrt x = \sqrt x \times \dfrac{{\sqrt x }}{{\sqrt x }} = \dfrac{{\sqrt {{x^2}} }}{{\sqrt x }} = \dfrac{x}{{\sqrt x }} $
So here $ \sqrt x $ is the rationalizing factor.
Now given polynomial is $ \sqrt[5]{{{a^2}{b^3}{c^4}}} $
Now as we see that it is 5th root of $ {a^2}{b^3}{c^4} $ so we have to multiply and divide by 5th root of $ {a^3}{b^2}c $ such that its numerator root is canceled out.
\[ \Rightarrow \sqrt[5]{{{a^2}{b^3}{c^4}}} = \sqrt[5]{{{a^2}{b^3}{c^4}}} \times \dfrac{{\sqrt[5]{{{a^3}{b^2}c}}}}{{\sqrt[5]{{{a^3}{b^2}c}}}} = \dfrac{{\sqrt[5]{{{a^5}{b^5}{c^5}}}}}{{\sqrt[5]{{{a^3}{b^2}c}}}} = \dfrac{{abc}}{{\sqrt[5]{{{a^3}{b^2}c}}}}\]
Therefore, rationalizing factor of $ \sqrt[5]{{{a^2}{b^3}{c^4}}} $ is \[\sqrt[5]{{{a^3}{b^2}c}}\].
So this is the required answer.
Hence option (A) is the correct answer.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall that the rationalizing factor is a term by which when we multiply and divide the given polynomial its numerator becomes a simple polynomial, so first check the root of the given polynomial then check the power of the individual variables then multiply and divide by the same root of the polynomial and inside the root the power of the variable is such that the sum of the power of the same variables is equal to the root number such that variables can easily come out from the given root.
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