Problem 32
Question
Simplify. $$ (6-z)(6+z) $$
Step-by-Step Solution
Verified Answer
The simplified form is \(36 - z^2\).
1Step 1: Understand the Expression
We have the expression \((6-z)(6+z)\). This resembles the form \((a-b)(a+b)\), which is a pattern known in algebra as the difference of squares.
2Step 2: Apply the Difference of Squares Formula
The difference of squares formula states: \((a-b)(a+b) = a^2 - b^2\). Here, \(a = 6\) and \(b = z\).
3Step 3: Substitute into the Formula
Substitute \(a = 6\) and \(b = z\) into the formula: \(a^2 - b^2 = 6^2 - z^2\).
4Step 4: Calculate the Squares
Calculate \(6^2 = 36\) and keep \(z^2\) as is, because it is in terms of a variable, resulting in \(36 - z^2\).
Key Concepts
Algebraic ExpressionsSimplifying ExpressionsPolynomial Identities
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and arithmetic operations, such as addition, subtraction, multiplication, and division. In the original exercise, we have two simple algebraic expressions, \(6 - z\) and \(6 + z\), multiplied together.
These expressions represent a particular type of binomial, which consists of two terms. Let's break down some essential components of algebraic expressions:
These expressions represent a particular type of binomial, which consists of two terms. Let's break down some essential components of algebraic expressions:
- Constants: Known numbers within the expression, such as 6 in our example.
- Variables: Symbols that represent unknown values, in this case, \(z\).
- Coefficients: Numbers multiplied by variables (here, the coefficient of \(z\) is -1 in \(6 - z\) and +1 in \(6 + z\)).
Simplifying Expressions
When we simplify expressions, we aim to reduce them to their most basic form while maintaining their original value. In the solved exercise, we start with the algebraic expression \((6-z)(6+z)\). The goal is to simplify this into a single, more concise expression.
One effective strategy is using special algebraic patterns, like the difference of squares. By recognizing that our expression fits this pattern as \((a-b)(a+b)\), we can apply the formula \(a^2-b^2\), where \(a = 6\) and \(b = z\).
The simplification involves these steps:
One effective strategy is using special algebraic patterns, like the difference of squares. By recognizing that our expression fits this pattern as \((a-b)(a+b)\), we can apply the formula \(a^2-b^2\), where \(a = 6\) and \(b = z\).
The simplification involves these steps:
- Identify the algebraic pattern.
- Substitute the values for \(a\) and \(b\).
- Perform the necessary arithmetic operations to achieve the simplest form, as in \(36 - z^2\).
Polynomial Identities
A polynomial identity is an equality that holds true for any values of the involved variables. The difference of squares, \((a-b)(a+b)=a^2-b^2\), provides a clear example of such an identity utilized in the original exercise. These identities help us transform products of expressions into simpler forms.
Polynomial identities are especially powerful tools because:
Polynomial identities are especially powerful tools because:
- They apply universally, no matter the numbers or variables involved.
- They enable swift computation, bypassing lengthy multiplication steps.
- They allow the transformation of complex expressions into more manageable forms.
Other exercises in this chapter
Problem 32
Find all of the zeros of each function. \(h(x)=x^{4}-15 x^{3}+70 x^{2}-70 x-156\)
View solution Problem 32
Solve each equation. $$ x^{4}+6 x^{2}-27=0 $$
View solution Problem 32
Simplify. $$ \frac{x^{4}+x^{2}-3 x+5}{x^{2}+2} $$
View solution Problem 32
Simplify. Assume that no variable equals 0. $$ \frac{12 x^{-3} y^{-2} z^{-8}}{30 x^{-6} y^{-4} z^{-1}} $$
View solution