Problem 38 - Precalculus Mathematics for Calculus

Question

Simplify each expression. (a) \(\frac{z^{2} z^{4}}{z^{3} z^{-1}}\) (b) \(\left(2 y^{2}\right)^{3}\) (c) \((8 x)^{2}\)

Step-by-Step Solution

Answer

(a) \( z^4 \), (b) \( 8y^6 \), (c) \( 64x^2 \).

Step 1: Simplify numerator and denominator in part (a)

In this step, simplify the expression by adding the exponents for both the numerator and the denominator. The original expression is \( \frac{z^2 z^4}{z^3 z^{-1}} \).Combine the terms in the numerator: \( z^2 \times z^4 = z^{2+4} = z^6 \).Combine the terms in the denominator:\( z^3 \times z^{-1} = z^{3 + (-1)} = z^{2} \).

Step 2: Divide powers of z in part (a)

Now divide the powers of \(z\) for expression \( \frac{z^6}{z^2} = z^{6-2} = z^4 \).Thus the simplified form is \( z^4 \).

Step 3: Expand the power in part (b)

The expression given is \( (2y^2)^3 \).First, apply the power to the constant and the variable:- Apply the cube to \( 2 \): \( 2^3 = 8 \).- Apply the cube to \( y^2 \) by multiplying the exponents: \((y^2)^3 = y^{2 \times 3} = y^6 \).

Step 4: Simplify expression in part (b)

Multiply the results from Step 3 to get the simplified expression: \( (2y^2)^3 = 8y^6 \).

Step 5: Expand the power and simplify in part (c)

For the expression \( (8x)^2 \), apply the square to both the constant and the variable part:- Square \(8\): \(8^2 = 64\).- Square \(x\): \((x)^2 = x^2\).Thus, the simplified form of \( (8x)^2 \) is \( 64x^2 \).

Key Concepts

Simplification of Expressions

Simplifying expressions is a process where we make expressions easier to read and work with by reducing them to their simplest form.
In the world of mathematics, especially with algebra, you can often simplify expressions to make them clearer and easier to solve.

• Start by identifying like terms or components that can be combined.
• Look for common bases when dealing with exponents, as they can often be simplified by using the rules of exponents.
• In fractions, divide the numerator by the denominator if they share the same base.

An example of simplification is taking an expression like \( \frac{z^{6}}{z^{2}} \) and simplifying it to \( z^{4} \). This is done by subtracting the exponent in the denominator from that in the numerator. Simplifying is essential because it makes complex expressions manageable and often paves the way for further solving.

Laws of Exponents

The laws of exponents are essential tools that simplify the multiplication and division of both numbers and variables with powers.
Understanding these laws allows for smoother calculations.

• **Product of Powers Law:** When multiplying like bases, add the exponents, like this: \( x^a \times x^b = x^{a+b} \).
• **Quotient of Powers Law:** When dividing like bases, subtract the exponents: \( \frac{x^a}{x^b} = x^{a-b} \).
• **Power of a Power Law:** To raise a power to another power, multiply the exponents: \( (x^a)^b = x^{a \times b} \).
• **Zero Exponent Law:** Any base (except zero) raised to the zero power is 1, such as \( x^0 = 1 \).

These laws are the cornerstone tools for simplifying and solving expressions that involve exponents, just like in the problem: \( \frac{z^2 z^4}{z^3 z^{-1}} \). Applying the relevant laws straightforwardly solves the expressions.

Power of a Product

This concept applies when you have a product of numbers and/or variables raised to an exponent.
Each number, coefficient, and variable within the parentheses must be raised to the power individually.

• This is described by the rule: \( (xy)^a = x^a y^a \).
• It's helpful to break down each component separately, allowing for easier computation and simplified results.

Consider \( (2y^2)^3 \): Instead of calculating the expression directly, raise 2 and \( y^2 \) separately to the third power, resulting in: \( 2^3 = 8 \) and \( (y^2)^3 = y^6 \). Multiplying these results gives \( 8y^6 \). By breaking it down, it streamlines and clarifies the process, making it more intuitive.

Multiplication of Exponents

When dealing with expressions raised to another power, the multiplication of exponents plays an integral role, especially in the power of a power situation.
Mastery of this concept involves knowing when to apply multiplication within the context of exponents.

• When you have an exponent outside parentheses, apply it to every component inside: \( (x^a)^b \) becomes \( x^{a \times b} \).
• This rule helps simplify more complex expressions efficiently.

For instance, in \( (8x)^2 \), independently apply the exponent to both 8 and x. This results in: \( 8^2 = 64 \) and \( x^2 \), giving a final simplified expression of \( 64x^2 \). By evenly distributing the power across all parts of the expression, clarity and simplicity are achieved.