Question

Find the quotient. $$ \frac{32 x^3 y}{y^8} \div \frac{y^7}{8 x^4} $$

Short answer

The quotient of the given division expression is \( \frac{256 x^7}{y^7} \).

Step-by-step solution

Rewrite the term as multiplication

Rewrite the given expression as multiplication of fractions instead of division, by flipping the second fraction. That is:\[\frac{32 x^3 y^1}{y^8} \times \frac{8 x^4}{y^7}\]

Cancel similar terms in the numerator and the denominator

Next, we make use of the rule of exponents, \(a^m/a^n = a^(m-n)\), by subtracting the powers of similar terms in the numerator and denominator. The goal is to simplify the expression.\[\frac{32 x^3 y^1}{y^8} \times \frac{8 x^4}{y^7} = \frac{32 \times 8 \times x^{(3+4)} \times y^{(1-7)}}{y^{8-7}} \]

Simplify further

Next, perform ordinary multiplication and addition or subtraction of the exponents according to the operations set in the previous step. This will give you:\[\frac{256 x^7 y^{-6}}{y^1}\]

Final simplification

Simplify the expression further by applying the rule of exponents, \(a^m/a^n = a^(m-n)\), one more time to obtain the simplest form.\[\frac{256 x^7 y^{-6}}{y^1} = \frac{256 x^7}{y^{1+6}} = \frac{256 x^7}{y^7}\]

Key concepts

Division of Polynomials

When working with algebraic fractions that involve division of polynomials, it's crucial to understand the process can be transformed into multiplication. This strategy simplifies the operation, and it's executed by taking the reciprocal (also known as multiplicative inverse) of the divisor, and then multiplying it by the dividend.

Let's take for instance our original problem where we need to divide \(\frac{32 x^3 y}{y^8} \) by \(\frac{y^7}{8 x^4} \). The first step is to flip the second fraction, changing the division into multiplication. The general principle here is that dividing by a number is the same as multiplying by its inverse. Thus, \(\frac{32 x^3 y}{y^8} \div \frac{y^7}{8 x^4} \) becomes \(\frac{32 x^3 y}{y^8} \times \frac{8 x^4}{y^7}\).

Once you've rewritten the problem as multiplication, you'll see that it becomes easier to simplify. This is because you can directly compare the numerators with the denominators and cancel out any like terms.

Exponent Rules

Grasping the exponent rules in algebra is vital for simplifying expressions with powers. Remember, when you multiply terms with the same base, you add their exponents. Conversely, dividing terms with the same base allows you to subtract the exponents of the divisor from the dividend.

In our example, we're simplifying \(\frac{32 x^3 y^1}{y^8} \times \frac{8 x^4}{y^7}\). The rules tell us to add exponents when multiplying like bases, so for the \(x\) terms: \(x^3 \times x^4 = x^{3+4} = x^7\). Similarly, we subtract exponents when dividing like bases, such as: \(y^1/y^8 = y^{1-8}\). These rules are the shortcut to streamline the problem without expanding terms into long, unwieldy expressions.

Another key rule to remember is dealing with negative exponents, such as \(y^{-6}\). The negative exponent tells you to take the reciprocal of the base, so \(y^{-6} = \frac{1}{y^6}\). Therefore, to simplify our final expression \(\frac{256 x^7 y^{-6}}{y^1}\), we apply this understanding to reach the final answer.

Simplifying Expressions

Simplifying expressions is an essential skill in algebra. It involves reducing expressions to their simplest form, making them more understandable and easier to work with. This can include combining like terms, factoring, expanding, and reducing fractions.

In our exercise, after applying the exponent rules, we're left with \(\frac{256 x^7 y^{-6}}{y^1}\), which still needs to be simplified. To do this, we look for any like terms and any opportunity to use the rules of exponents to reduce the expression further. In this case, we have a negative exponent in the numerator and a positive exponent in the denominator, which can be combined to yield \(\frac{256 x^7}{y^{1+6}}\) or \(\frac{256 x^7}{y^7}\).

This concludes the simplification process, and it illustrates how breaking down the problem into manageable parts by using rules for division of polynomials, exponent rules, and general principles for simplifying expressions can lead to a neat and understandable solution.