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Q.

The rationalizing factor of $\sqrt[5]{a^{2} b^{3} c^{4}}$ is:

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a

\sqrt[5]{a^{3} b^{2} c}

b

\sqrt[5]{a^{3} bc}

c

\sqrt[5]{a^{3} b^{2} c^{5}}

d

\sqrt[5]{a^{3} b^{6} c}

answer is A.

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Detailed Solution

Concept- Rationalizing factor of the polynomial here, is a technique such that the

n^{t h}

root polynomial is converted into a simple polynomial i.e., without

n^{t h}

root or any other root by multiplying and dividing by the polynomial such that the numerator root is cancelled out.
Given

\sqrt[5]{a^{2} b^{3} c^{4}}

,
As, it is

5^{t h}

root of

a^{2} b^{3} c^{4}

,so we have to multiply and divide by

5^{t h}

root of

a^{3} b^{2} c^{2}

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 \frac{\sqrt[5]{a^{3} b^{2} c}}{\sqrt[5]{a^{3} b^{2} c}}

\Rightarrow \sqrt[5]{a^{2} b^{3} c^{4}} = \frac{\sqrt[5]{a^{5} b^{5} c^{5}}}{\sqrt[5]{a^{3} b^{2} c}}

\Rightarrow \sqrt[5]{a^{2} b^{3} c^{4}} = \frac{abc}{\sqrt[5]{a^{3} b^{2} c}}

Therefore, the rationalizing factor of

\sqrt[5]{a^{2} b^{3} c^{4}}

is

\sqrt[5]{a^{3} b^{2} c}

Hence, the correct answer is option 1.

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