Problem 57 - Intermediate Algebra

Question

Divide, and then simplify, if possible. See Example 7. $$ \frac{a^{2}-18 a+81}{a^{40}} \div \frac{(a-9)^{3}}{a^{37}} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{1}{(a - 9)a^{3}} \).

Step 1: Write the Division as Multiplication

When dividing by a fraction, you can multiply by its reciprocal. The expression \( \frac{a^{2}-18a+81}{a^{40}} \div \frac{(a-9)^3}{a^{37}} \) becomes \( \frac{a^{2}-18a+81}{a^{40}} \times \frac{a^{37}}{(a-9)^3} \).

Step 2: Simplify Inside the Numerator

Factor the quadratic expression \( a^2 - 18a + 81 \) in the numerator. Notice that \( a^2 - 18a + 81 = (a - 9)^2 \). Now the expression becomes \( \frac{(a-9)^2}{a^{40}} \times \frac{a^{37}}{(a-9)^3} \).

Step 3: Multiply the Fractions

Multiply the two fractions: \( \frac{(a-9)^2 \cdot a^{37}}{a^{40} \cdot (a-9)^3} \).

Step 4: Simplify the Expression

Combine and simplify terms: The \((a-9)^2\) on the numerator will cancel out two of the \((a-9)\)'s in the denominator, leaving \( a-9 \) in the denominator. For the exponents of \(a\), \( a^{37} \) over \( a^{40} \) simplifies to \( \frac{1}{a^{3}} \). The expression is now \( \frac{1}{(a - 9)a^{3}} \).

Key Concepts

Division of Fractions

When dealing with division involving fractions, a handy trick to remember is to multiply by the reciprocal. This method transforms the problem into something more familiar and less daunting. For instance, consider the original problem of dividing \( \frac{a^{2}-18 a+81}{a^{40}} \div \frac{(a-9)^{3}}{a^{37}} \). Instead of directly dividing these fractions, swap the numerator and denominator of the second fraction—this is its reciprocal.

• Original: \( \frac{a^{2}-18a+81}{a^{40}} \div \frac{(a-9)^3}{a^{37}} \)
• Reciprocal form: \( \frac{a^{2}-18a+81}{a^{40}} \times \frac{a^{37}}{(a-9)^3} \)

By converting division into multiplication, we align fractions in such a way that they can be managed more easily. Afterwards, it's time to look into simplifying the expression which involves factoring and reducing.

Factoring Quadratics

Factoring quadratics is a key tool in simplifying expressions, particularly when they form part of a fraction. In this scenario, our goal is to break down the quadratic expression in the numerator: \( a^2 - 18a + 81 \). Upon inspection, this expression is a perfect square trinomial. This means it can be rewritten as the square of a binomial.

• Quadratic: \( a^2 - 18a + 81 \)
• Factored form: \( (a - 9)^2 \)

Recognizing perfect square trinomials and being able to rewrite them is an instrumental skill. This factorization will play a crucial role when simplifying expressions, particularly as we move forward to multiply fractions and reduce common terms.

Simplifying Expressions

Simplifying expressions is a process that often follows factoring and involves reducing fractions to their simplest form. After factoring the quadratic and setting up the multiplication, we have:

• Expression: \( \frac{(a-9)^2}{a^{40}} \times \frac{a^{37}}{(a-9)^3} \)

To simplify, let's first cancel out any common terms. Here, the \((a-9)^2\) in the numerator matches with part of \((a-9)^3\) in the denominator. Reducing these terms simplifies the expression. Additionally, consider the powers of \(a\):

• \(a^{37} \div a^{40} \) results in \(a^{37-40} = \frac{1}{a^3}\).

The simplification process is like balancing an equation, striving for the simplest possible form which, in this case, results in \( \frac{1}{(a - 9)a^{3}} \).

Multiplication of Fractions

Once we convert the division into multiplication and factor any necessary expressions, it's time to multiply the fractions. This process is straightforward:

• Step 1: Multiply the numerators together.
  • Numerators: \((a-9)^2 \times a^{37}\)


• Step 2: Multiply the denominators.
  • Denominators: \(a^{40} \times (a-9)^3\)

The key after multiplication is identifying and simplifying common terms, like \((a-9)^2\) over \((a-9)^3\), or balancing the powers of \(a\). Remember, the essence of these steps is to make the expression as simple as possible, highlighting the power of multiplication in fraction operations!