Problem 72 - Algebra 2

Question

Which expression is a factor of \(x^{4}-18 x^{2}+81 ?\) $$ \begin{array}{llll}{\text { A. } x^{2}-9} & {\text { B. } x^{2}+6 x-9} & {\text { C. } x^{2}-6 x-9} & {\text { D. } x^{2}+9}\end{array} $$

Step-by-Step Solution

Answer

The expression that is a factor of \(x^{4}-18x^{2}+81\) is \(x^{2}-9\), which is option A.

Step 1: Recognize a Pattern

Notice that the given polynomial, \(x^{4}-18x^{2}+81\), resembles the structure of a squared binomial, since \(81 = 9^2\) and \(18x^{2} = 2 \times 9 \times x^{2}\). We can rewrite the polynomial as \((x^{2})^{2} - 2 \times 9 \times x^{2} + 9^{2}\).

Step 2: Apply the Difference of Squares Formula

We look for an expression that factors into a perfect square trinomial. The difference of squares formula is \(a^{2} - b^{2} = (a - b)(a + b)\). Given that \(x^{4} = (x^{2})^{2}\) and \(81 = 9^{2}\), we can consider the trinomial as the square of \(x^{2} - 9\). Thus, \((x^{2} - 9)^{2} = x^{4} - 18x^{2} + 81\).

Step 3: Identify the Correct Factor

Now that we know \((x^{2}-9)^{2} = x^{4} - 18x^{2} + 81\), the factor of the original expression \(x^{4}-18x^{2}+81\) is one of the two identical factors of the squared expression, which is \(x^{2} - 9\). Therefore, option A is the correct factor.

Key Concepts

Difference of Squares

Understanding the difference of squares is a fundamental concept in algebra that can significantly simplify factoring certain types of polynomials. The difference of squares refers to an expression in the form \( a^2 - b^2 \). This expression can always be factored into \( (a - b)(a + b) \).

Let's consider why this works. If we multiply \( (a - b) \) by \( (a + b) \), we end up with \( a^2 - ab + ab - b^2 \). You'll notice that the middle terms, \( -ab \) and \( ab \), cancel each other out, leaving us with just \( a^2 - b^2 \).

In the context of our exercise, we recognized that \( x^4 \) is the square of \( x^2 \), and \( 81 \) is the square of \( 9 \). Therefore, we could apply the difference of squares technique to factor our polynomial, which further simplifies the process of finding its factors.

Perfect Square Trinomial

When dealing with quadratic equations and expressions, the perfect square trinomial is another pattern to be aware of. A perfect square trinomial looks like \( a^2 \text{±} 2ab + b^2 \), where the terms are the squares of binomials \( (a \text{±} b)^2 \). This pattern arises when we square a binomial expression.When you square \( a + b \), for example, you get \( a^2 + 2ab + b^2 \). Similarly, squaring \( a - b \) results in \( a^2 - 2ab + b^2 \). The middle term is always twice the product of the two terms being squared.In the exercise, we noticed that the given polynomial fits the structure of a squared binomial. Recognizing this pattern allowed us to rewrite the polynomial in the form of a perfect square trinomial, \( (x^2 - 9)^2 \). Consequently, being able to identify a perfect square trinomial can be incredibly useful and is a key step in factoring and solving more complex algebraic expressions.

Algebraic Expressions

Algebraic expressions are the building blocks of algebra and consist of numbers, variables, and arithmetic operations. They can be as simple as \( x + 5 \) or as complex as \( 4x^3 - 5x^2 + 7x - 8 \). An important skill in algebra is being able to manipulate these expressions to make solving problems easier, whether by simplifying, expanding, factoring, or solving for a variable.Factoring is a crucial technique for simplifying algebraic expressions, especially polynomials. It involves breaking down a complex expression into simpler parts or 'factors' that, when multiplied together, yield the original expression. One commonly sought-after form of factoring is looking for expressions that can be simplified as products of binomials or trinomials, as we did in our previous steps with the given polynomial.In practice, students should be equipped with recognizing patterns, applying relevant formulas, and strategically manipulating these expressions to make other operations more manageable. This problem highlighted such an approach, utilizing both the recognition of a perfect square trinomial and the difference of squares to find our factor.