Answer
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Hint: To solve this question, we will factorize the given monomials $7{{x}^{2}}y{{z}^{4}},21{{x}^{2}}{{y}^{5}}{{z}^{3}}$ and take out the common terms of their factorization. This common term will give us the greatest common divisor of $7{{x}^{2}}y{{z}^{4}},21{{x}^{2}}{{y}^{5}}{{z}^{3}}$.
Complete step by step solution:
It is given in the question that we have to find the greatest common divisor of: $7{{x}^{2}}y{{z}^{4}},21{{x}^{2}}{{y}^{5}}{{z}^{3}}$. Now, if we factorize the given monomials, $7{{x}^{2}}y{{z}^{4}},21{{x}^{2}}{{y}^{5}}{{z}^{3}}$ we will get as follows,
$\begin{align}
& 7{{x}^{2}}y{{z}^{4}}=7\times x\times x\times y\times z\times z\times z\times z \\
& 21{{x}^{2}}{{y}^{5}}{{z}^{3}}=7\times 3\times x\times x\times y\times y\times y\times y\times y\times z\times z\times z \\
\end{align}$
Now, if we take out the common terms from the factorization of \[7{{x}^{2}}y{{z}^{4}}\] and \[21{{x}^{2}}{{y}^{5}}{{z}^{3}}\], we will get
$7\times x\times x\times y\times z\times z\times z$ as common.
We can write the above expression as follows, $7{{x}^{2}}y{{z}^{3}}$.
Hence, we get that the greatest common divisor of the term \[7{{x}^{2}}y{{z}^{4}}\] and \[21{{x}^{2}}{{y}^{5}}{{z}^{3}}\] is $7{{x}^{2}}y{{z}^{3}}$.
Therefore, option (a) is the correct answer.
Note: There is a possibility that many students will make a mistake by multiplying the given monomials \[7{{x}^{2}}y{{z}^{4}}\] and \[21{{x}^{2}}{{y}^{5}}{{z}^{3}}\] and then they will write their answer, that is, the greatest common divisor is the product of the given monomials, that are, \[7{{x}^{2}}y{{z}^{4}}\] and \[21{{x}^{2}}{{y}^{5}}{{z}^{3}}\], but this is not correct. So, the students must note that the greatest common divisor is always less than or equal to the smaller number given in the list so, if the students take the product of the given monomials, that are, \[7{{x}^{2}}y{{z}^{4}}\] and \[21{{x}^{2}}{{y}^{5}}{{z}^{3}}\], it will definitely be greater than \[7{{x}^{2}}y{{z}^{4}}\], hence it is wrong to take the product of the monomials.
Complete step by step solution:
It is given in the question that we have to find the greatest common divisor of: $7{{x}^{2}}y{{z}^{4}},21{{x}^{2}}{{y}^{5}}{{z}^{3}}$. Now, if we factorize the given monomials, $7{{x}^{2}}y{{z}^{4}},21{{x}^{2}}{{y}^{5}}{{z}^{3}}$ we will get as follows,
$\begin{align}
& 7{{x}^{2}}y{{z}^{4}}=7\times x\times x\times y\times z\times z\times z\times z \\
& 21{{x}^{2}}{{y}^{5}}{{z}^{3}}=7\times 3\times x\times x\times y\times y\times y\times y\times y\times z\times z\times z \\
\end{align}$
Now, if we take out the common terms from the factorization of \[7{{x}^{2}}y{{z}^{4}}\] and \[21{{x}^{2}}{{y}^{5}}{{z}^{3}}\], we will get
$7\times x\times x\times y\times z\times z\times z$ as common.
We can write the above expression as follows, $7{{x}^{2}}y{{z}^{3}}$.
Hence, we get that the greatest common divisor of the term \[7{{x}^{2}}y{{z}^{4}}\] and \[21{{x}^{2}}{{y}^{5}}{{z}^{3}}\] is $7{{x}^{2}}y{{z}^{3}}$.
Therefore, option (a) is the correct answer.
Note: There is a possibility that many students will make a mistake by multiplying the given monomials \[7{{x}^{2}}y{{z}^{4}}\] and \[21{{x}^{2}}{{y}^{5}}{{z}^{3}}\] and then they will write their answer, that is, the greatest common divisor is the product of the given monomials, that are, \[7{{x}^{2}}y{{z}^{4}}\] and \[21{{x}^{2}}{{y}^{5}}{{z}^{3}}\], but this is not correct. So, the students must note that the greatest common divisor is always less than or equal to the smaller number given in the list so, if the students take the product of the given monomials, that are, \[7{{x}^{2}}y{{z}^{4}}\] and \[21{{x}^{2}}{{y}^{5}}{{z}^{3}}\], it will definitely be greater than \[7{{x}^{2}}y{{z}^{4}}\], hence it is wrong to take the product of the monomials.
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