Problem 52 - Algebra A Combined Function

Question

Factor each completely. $$ x^{2}-18 x+81-y^{2} $$

Step-by-Step Solution

Answer

The expression factors to \((x-9-y)(x-9+y)\).

Step 1: Identify the Expression

The given expression is a combination of a quadratic and a term involving \(y\). It can be rewritten as: \[x^2 - 18x + 81 - y^2\].

Step 2: Recognize the Form of the Quadratic

Notice that the quadratic part \(x^2 - 18x + 81\) can potentially be a perfect square trinomial. Let's check if this is the case.

Step 3: Verify if the Quadratic is a Perfect Square

A perfect square trinomial can be expressed as \((x-a)^2\). For \(x^2 - 18x + 81\), check if 81 is the square of half of the coefficient of \(x\): - Half of \(-18\) is \(-9\).- \((-9)^2 = 81\).Therefore, \(x^2 - 18x + 81\) is a perfect square: \((x-9)^2\).

Step 4: Rewrite the Expression

Using the result from Step 3, rewrite the expression as: \[(x-9)^2 - y^2\].

Step 5: Recognize the Difference of Squares

The expression \((x-9)^2 - y^2\) is a difference of squares. It can be factored using the formula: \(a^2 - b^2 = (a-b)(a+b)\).

Step 6: Factor using the Difference of Squares

Let \(a = (x-9)\) and \(b = y\). Then factor the expression:\[(x-9-y)(x-9+y)\].

Step 7: Present the Complete Factorization

The completely factored form of \(x^2 - 18x + 81 - y^2\) is: \((x-9-y)(x-9+y)\).

Key Concepts

Perfect Square Trinomial

In algebra, a **perfect square trinomial** is a quadratic expression that can be written as the square of a binomial. Identifying these trinomials is essential when factoring expressions because they can simplify into a neat square form. Let's break down what makes a trinomial a perfect square:

• A trinomial is an expression of the form \(ax^2 + bx + c\).
• For it to be a perfect square, there should exist a value \(a\), such that the trinomial can be expressed as \((a - b)^2\) or \((a + b)^2\).
• The **key property** here is that the middle term \(b\) should equal \(2ab\), and the last term \(c\) should equal \(b^2\).

For example, the expression \(x^2 - 18x + 81\) can be seen as \((x - 9)^2\). Notice that:

• The coefficient of \(x\) is \(-18\), and half of this number, squared, gives us \(81\).
• Therefore, it fits the perfect square trinomial form, hence allowing it to be simplified into a binomial squared.

Understanding this opens the door to easier factorization and streamlines solutions.

Difference of Squares

The **difference of squares** formula is another cornerstone of algebraic factorization. It deals with expressions of the form \(a^2 - b^2\), allowing them to be rewritten into a product of their square roots. Here’s how it works:

• The expression must strictly be the difference (subtraction) of two perfect squares.
• It can be then factored into \((a - b)(a + b)\).

This is handy in the given problem where after recognizing a perfect square trinomial as \((x - 9)^2\), the expression becomes \((x - 9)^2 - y^2\), a classic difference of squares:

• Let \(a = (x - 9)\) and \(b = y\).
• The expression can be factored into \((x - 9 - y)(x - 9 + y)\).

Knowing how to apply the difference of squares allows quick and accurate factorization of specific quadratic expressions, saving time and reducing errors.

Quadratic Expressions

**Quadratic expressions** are fundamental in algebra and appear in countless mathematical problems. A quadratic expression is any polynomial expression of the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is a variable.

• The highest degree of \(x\) in a quadratic expression is 2.
• It frequently arises in equations that describe parabolic graphs and physical scenarios.

When factoring quadratic expressions, the main goal is to express them as a product of simpler binomials. Understanding how to break them down involves:

• Recognizing special patterns like perfect square trinomials (learned earlier) and applying the difference of squares when applicable.
• In cases where these methods don't apply, alternative techniques like the quadratic formula or completing the square might be applied.

Quadratics are essential not only because they are common, but also because they represent the foundation upon which more complex algebraic operations are built. Mastery over their manipulation is a key step in advancing your algebra skills.