Problem 8 - Elementary Algebra

Question

Simplify each square root. $$ \sqrt{180 m^{4} n^{15}(a-12)^{15}} $$

Step-by-Step Solution

Answer

Question: Simplify the given square root expression: $$\sqrt{180 m^{4} n^{15}(a-12)^{15}}$$ Answer: The simplified expression is: $$6 m^2 n^7 \sqrt{5 n (a-12)^{15}}$$

Step 1: Factor the numeric coefficient and exponents

We need to factor the given expression, especially the coefficient and exponents, to find any perfect squares that can be simplified. $$ \sqrt{180 m^{4} n^{15}(a-12)^{15}} = \sqrt{2^2 \cdot 3^2 \cdot 5 m^{4} n^{2 \cdot 7 + 1}(a-12)^{15}} $$

Step 2: Take the square root of the perfect squares

Now that we have factored out the expression, we can take the square root of all the perfect squares in the expression. Perfect squares in this case are: \(2^2\), \(3^2\), \(m^4\), \(n^{2 \cdot 7}\), and \((a-12)^{2 \cdot 7}\). The square root of each perfect square is equal to half of its exponent. $$ \sqrt{2^2 \cdot 3^2 \cdot 5 m^{4} n^{2 \cdot 7 + 1}(a-12)^{15}} = 2 \cdot 3 m^2 n^7 \sqrt{5 n (a-12)^{15}} $$

Step 3: Rewrite the result

We now have the simplified expression after taking the square root of all the perfect squares in the original expression. $$ \sqrt{180 m^{4} n^{15}(a-12)^{15}} = 6 m^2 n^7 \sqrt{5 n (a-12)^{15}} $$

Key Concepts

Perfect Squares

A perfect square is a number or expression that can be expressed as the square of an integer or a polynomial. Recognizing perfect squares is crucial when simplifying square roots, as it allows us to rewrite certain parts of the expression under the square root as a simpler term.
For instance, numbers like 4, 9, and 16 are perfect squares because they can be written as:

• 4 = 2^2
• 9 = 3^2
• 16 = 4^2

Similarly, algebraic terms like \( m^4 \) are perfect squares since \( m^4 = (m^2)^2 \). Recognizing and utilizing these properties allows us to simplify complex expressions under a square root.
When dealing with exponents, perfect squares are exponents that are multiples of 2, such as \( n^{14} \), since it can be rewritten as \((n^7)^2\). This makes the simplification process quite straightforward as you just take the square of the exponent, essentially halving it.

Exponent Rules

Understanding exponent rules is vital when working with terms under a square root, especially with variables. Exponent rules help us factor expressions and identify perfect squares. Some fundamental exponent rules include:

• \( x^a \cdot x^b = x^{a+b} \)
• \( (x^a)^b = x^{a\cdot b} \)
• \( x^{-a} = \frac{1}{x^a} \)

In the given problem, these rules allow us to transform expressions like \( n^{15} \) into a format that reveals perfect squares. By expressing \( n^{15} \) as \( n^{2\cdot 7 + 1} \), we can easily separate it into \((n^7)^2\cdot n^1\), where \((n^7)^2\) is a perfect square.
This transformation is essential in simplifying square roots, as it helps reduce the complexity and clarify which parts of the expression can be simplified further.

Factoring Expressions

Factoring expressions is the process of breaking down a complex expression into a product of simpler terms. Factoring is particularly useful when identifying perfect squares and simplifying square roots. For example, when factoring \(180\), we break it down into its prime factors: \(2^2 \cdot 3^2 \cdot 5\).
Using factoring, we identify any perfect square components, such as \(2^2\) and \(3^2\), which can be simplified when under a square root.
This approach also extends to algebraic expressions. Take \(m^4\), for example, which can be seen as \((m^2)^2\), thus simplifying to \(m^2\). Similarly, expressions involving binomials like \((a-12)^{15}\) are intentionally grouped to show the exponential factor is odd, guiding which terms remain after simplifying.
Factoring is a necessary skill in algebra that unveils smaller, more manageable elements in an expression, aiding the simplification process, particularly for square roots.