Problem 55 - Algebra A Combined Function

Question

Multiply. $$ (9 z+5)(9 z-5) $$

Step-by-Step Solution

Answer

\(81z^2 - 25\)

Step 1: Identify the Expression Type

The expression \((9z+5)(9z-5)\) is a product of conjugates. This means it is in the form of \((a+b)(a-b)\), where \(a = 9z\) and \(b = 5\).

Step 2: Apply the Difference of Squares Formula

For the conjugates \((a+b)(a-b)\), we use the difference of squares formula: \((a+b)(a-b) = a^2 - b^2\). Substitute \(a = 9z\) and \(b = 5\) into the formula.

Step 3: Calculate \(a^2\)

Substitute \(a = 9z\) into \(a^2\). Therefore, \(a^2 = (9z)^2 = 81z^2\).

Step 4: Calculate \(b^2\)

Substitute \(b = 5\) into \(b^2\). Therefore, \(b^2 = 5^2 = 25\).

Step 5: Write Out the Final Expression

Using the difference of squares result \(a^2 - b^2\), substitute the values from Steps 3 and 4: \(81z^2 - 25\). This is your final answer.

Key Concepts

Difference of Squares

The difference of squares is a powerful algebraic concept. It allows us to simplify expressions that fit a specific pattern. The formula is given by \[ (a + b)(a - b) = a^2 - b^2 \] and applies when you have two conjugate binomials, such as \((a + b) \text{ and } (a - b)\). The magic of this pattern lies in the fact that multiplying these pairs always results in the difference of two squares.
For example, taking the exercise \( (9z + 5)(9z - 5) \):

• We identify \( a = 9z \) and \( b = 5 \).
• Using the formula, we get \( a^2 - b^2 \), which will give us the result \( 81z^2 - 25 \).

This formula is particularly helpful because it cuts down on the steps needed for traditional binomial multiplication.

Product of Conjugates

Conjugates in algebra refer to pairs of binomials where each term \( a + b \) and \( a - b \) have the same first term and opposite second terms. An important algebraic property is that when you multiply conjugates, you always get a difference of squares:

• Conjugates include pairs like \((x + y) \text{ and } (x - y)\).
• In this setup, the multiplication results in \( x^2 - y^2 \).

In our function \((9z + 5)(9z - 5)\),

• The terms \((9z) \text{ and } (5)\) serve as our \( a \text{ and } b\) respectively.
• This means the result is the square of \( 9z \) minus the square of \( 5 \), or \( 81z^2 - 25 \).

Recognizing and using conjugates can make binomial multiplication swift and efficient.

Polynomials

Polynomials are a core concept in algebra. They are expressions involving multiple terms constructed from variables (\( z \)) and coefficients. Often, these terms are raised to powers. A binomial is a type of polynomial with exactly two terms like \(9z + 5\) or \(9z - 5\).
They can be thought of as building blocks for more complex algebraic expressions. The multiplication of binomials often yields polynomials with terms combining the variables and their powers.
In our exercise,

• The resulting expression \(81z^2 - 25\) is a polynomial formed by a variable term squared and a constant term.
• This highlights the change from a two-term expression into a potentially simpler polynomial through multiplication.

The simplicity of polynomials often makes them easier to work with and understand in different algebraic scenarios.

Multiplication of Binomials

Multiplying binomials can seem challenging, but it becomes straightforward when using specific patterns. One such pattern is the difference of squares, which we use when multiplying conjugates. In general, for any two binomials \((a + b)\text{ and }(c + d)\),

• The multiplication process involves: \( a \times c, \) \( a \times d, \)\( b \times c,\) and \( b \times d.\)
• The result is a polynomial with up to four terms: \( ac, ad, bc, \text{ and } bd.\)

However, in our exercise with \( (9z + 5)(9z - 5) \), notice how elegantly the structure allows simplification into fewer terms.

• The special form leads directly to \( a^2 - b^2 \): \( 81z^2 - 25 \).

Understanding these patterns and practicing them can significantly ease the process of multiplying and simplifying binomials.