Find the square root of ${a^{2n + 2}}{b^{2n + 4}}{c^{2n + 6}}$.
$
A.{a^{n + 1}}{b^{n + 2}}{c^{n + 3}} \\
B.{a^{n + 2}}{b^{n + 1}}{c^{n + 3}} \\
C.{a^{n + 3}}{b^{n + 2}}{c^{n + 1}} \\
D.{\text{ all of the above}} \\
$
Answer
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Hint: When the number is multiplied by itself, it results in the square of a number. Alternatively, when a number is raised to the power of half, it results in the square root of a number. The square root of a number is denoted as \[\sqrt x \], and so, when the result is multiplied by itself, produces \[\sqrt x \times \sqrt x = x\] the original number. Here, in the question, a function of the polynomial has been given, so the easiest way to determine the square root of the given function is to simplify the expression of the function and calculate the square root of the simplified function by following the properties of the power.
Complete step by step solution:
The given function can be re-written as:
\[ {a^{2n + 2}}{b^{2n + 4}}{c^{2n + 6}} = {a^{2n}} \times {a^2} \times {b^{2n}} \times {b^4} \times {c^{2n}} \times {c^6} \\
= {\left( {abc} \right)^{2n}} \times {a^2} \times {b^4} \times {c^6} \\
= {\left( {abc} \right)^{2n}} \times {\left( {a{b^2}{c^3}} \right)^2} \\ \]
Now, taking the square root to both the sides of the equation:
\[ \sqrt {\left( {{a^{2n + 2}}{b^{2n + 4}}{c^{2n + 6}}} \right)} = \sqrt {{{\left( {abc} \right)}^{2n}} \times {{\left( {a{b^2}{c^3}} \right)}^2}} \\
= \sqrt {{{\left( {\left( {{{\left( {abc} \right)}^n}} \right) \times \left( {a{b^2}{c^3}} \right)} \right)}^2}} \\
= \left( {{{\left( {abc} \right)}^n} \times \left( {a{b^2}{c^3}} \right)} \right) \\
= {a^{n + 1}} \times {b^{n + 2}} \times {c^{n + 3}} \\ \]
Hence, the square root ${a^{2n + 2}}{b^{2n + 4}}{c^{2n + 6}}$ is ${a^{n + 1}}{b^{n + 2}}{c^{n + 3}}$.
Option A is correct.
Additional Information: One of the properties of powers is ${x^{n + 2}} = {x^n} \times {x^2}$.
Note: Square root of a number can be produced either by the long division method or by the prime factorization method. The numbers which have only two factors, i.e., 1 and itself, are termed as prime numbers. To find the square root of a number using the method of prime factorization first, we will have to find the prime factor of the number, and these numbers are grouped in pairs, which are the same, and then their product is found. For example, the prime factor of \[\left( c \right) = a \times a \times a \times b \times b \times a \times c \times c\] the number which is grouped in pair as \[\left( c \right) = \underline {a \times a} \times \underline {a \times a} \times \underline {b \times b} \times \underline {c \times c} = a \times a \times b \times c\]
Complete step by step solution:
The given function can be re-written as:
\[ {a^{2n + 2}}{b^{2n + 4}}{c^{2n + 6}} = {a^{2n}} \times {a^2} \times {b^{2n}} \times {b^4} \times {c^{2n}} \times {c^6} \\
= {\left( {abc} \right)^{2n}} \times {a^2} \times {b^4} \times {c^6} \\
= {\left( {abc} \right)^{2n}} \times {\left( {a{b^2}{c^3}} \right)^2} \\ \]
Now, taking the square root to both the sides of the equation:
\[ \sqrt {\left( {{a^{2n + 2}}{b^{2n + 4}}{c^{2n + 6}}} \right)} = \sqrt {{{\left( {abc} \right)}^{2n}} \times {{\left( {a{b^2}{c^3}} \right)}^2}} \\
= \sqrt {{{\left( {\left( {{{\left( {abc} \right)}^n}} \right) \times \left( {a{b^2}{c^3}} \right)} \right)}^2}} \\
= \left( {{{\left( {abc} \right)}^n} \times \left( {a{b^2}{c^3}} \right)} \right) \\
= {a^{n + 1}} \times {b^{n + 2}} \times {c^{n + 3}} \\ \]
Hence, the square root ${a^{2n + 2}}{b^{2n + 4}}{c^{2n + 6}}$ is ${a^{n + 1}}{b^{n + 2}}{c^{n + 3}}$.
Option A is correct.
Additional Information: One of the properties of powers is ${x^{n + 2}} = {x^n} \times {x^2}$.
Note: Square root of a number can be produced either by the long division method or by the prime factorization method. The numbers which have only two factors, i.e., 1 and itself, are termed as prime numbers. To find the square root of a number using the method of prime factorization first, we will have to find the prime factor of the number, and these numbers are grouped in pairs, which are the same, and then their product is found. For example, the prime factor of \[\left( c \right) = a \times a \times a \times b \times b \times a \times c \times c\] the number which is grouped in pair as \[\left( c \right) = \underline {a \times a} \times \underline {a \times a} \times \underline {b \times b} \times \underline {c \times c} = a \times a \times b \times c\]
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