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Chapter 5: Problem 60

Factor completely. See Examples 1 through \(9 .\) $$ x^{2}-18 x+81-y^{2} $$

Short Answer

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

Step by step solution

01

Identify the Type of Expression

The given expression is \( x^2 - 18x + 81 - y^2 \). Notice the structure: it comprises a quadratic trinomial \( x^2 - 18x + 81 \) and a \( y^2 \) term. This can be converted into a difference of squares.
02

Factor the Quadratic Trinomial

The expression \( x^2 - 18x + 81 \) is a perfect square trinomial. It can be rewritten as \( (x-9)^2 \) since \((x-9)(x-9) = x^2 - 18x + 81\). Check by expanding: \[(x-9)^2 = x^2 - 18x + 81.\]
03

Express as a Difference of Squares

Now substitute the quadratic trinomial with its factored form: \( (x-9)^2 - y^2 \). This expression is now in the form of a difference of squares: \[ (A)^2 - (B)^2, \] where \( A = (x-9) \) and \( B = y \).
04

Factor the Difference of Squares

The difference of squares \( (A)^2 - (B)^2 \) can be factored using the formula \[ A^2 - B^2 = (A + B)(A - B). \] So, \[(x-9)^2 - y^2 = ((x-9) + y)((x-9) - y).\]
05

Write the Final Factored Expression

Write down the factored form of the initial expression: \((x-9+y)(x-9-y)\). This is the completely factored form of the original expression \(x^2 - 18x + 81 - y^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Squares
The difference of squares is a crucial method in factoring polynomials. It applies when you have two terms, both squares, separated by a subtraction sign. Let's break it down further. The general form is \( a^2 - b^2 \), where both \( a \) and \( b \) are any expressions that can be squared.

To factor a difference of squares, use the identity: \( a^2 - b^2 = (a + b)(a - b) \). This formula is extremely useful because it simplifies expressions into a product of two binomials. The expression \( (x-9)^2 - y^2 \) fits this pattern.
  • Let \( A = x - 9 \), the first square.
  • Let \( B = y \), the second square.
By applying the identity directly, \( (x-9)^2 - y^2 \) becomes \((x-9 + y)(x-9 - y)\). This breaks down a complex polynomial into simpler, manageable parts.
Quadratic Trinomials
A quadratic trinomial is a polynomial expression with three terms, shaped generally like \( ax^2 + bx + c \). These are common in algebra and mastering their factoring is essential. The key is learning to transform them into binomials, like \((x - p)(x - q)\).

In this case, consider \( x^2 - 18x + 81 \). It looks intimidating, but it is designed as a "perfect square trinomial."
  • The perfect squares are important. Notice the squares: \( x^2 \) and \( 81 = 9^2 \).
  • The middle term, \(-18x\), is twice the product of \( x \) and \( 9 \). This confirms it is a perfect square trinomial.
Therefore, \( x^2 - 18x + 81 \) factors neatly into \((x - 9)^2\). Recognizing perfect square trinomials helps quickly reduce polynomials into simpler forms.
Perfect Square Trinomials
Perfect square trinomials are a special type of quadratic expressions. They arise from squaring a binomial, which results in a neat, predictable pattern. The general form for perfect square trinomials is \((a \,\pm b)^2 = a^2 \pm 2ab + b^2\).

In our example, \( x^2 - 18x + 81 \) is a perfect square trinomial. It perfectly matches the pattern \((x-9)^2 = x^2 - 18x + 81\). Here's why:
  • First term: \( x^2 \) is \( (x)^2 \).
  • Middle term: \(-18x\) is \(-2(x)(9)\), fitting the \(-2ab\) part of the pattern.
  • Last term: \( 81 \) is \( (9)^2 \).
Recognizing perfect square trinomials simplifies many algebraic problems, making calculations more straightforward and manageable. Once identified, these can immediately be factored into simpler binomial expressions like \((x-9)^2\).

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Most popular questions from this chapter

Factor each polynomial completely. See Examples 1 through 12. $$ 12 x^{2}-17 x+6 $$

Factor each polynomial completely. See Examples 1 through 12. $$ 20 x^{5}+54 x^{4}+10 x^{3} $$

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Find the value of \(c\) that makes each trinomial a perfect square trinomial. Factor \(x^{6}-1\) completely, using the following methods from this chapter. A. Factor the expression by treating it as the difference of two squares, \(\left(x^{3}\right)^{2}-1^{2}\) B. Factor the expression by treating it as the difference of two squares, \(\left(x^{3}\right)^{2}-1^{2}\) C. Factor the expression by treating it as the difference of two squares, \(\left(x^{3}\right)^{2}-1^{2}\)

The function \(f(x)=0.0007 x^{2}+0.24 x+7.98\) can be used to approximate the total cheese production in the United States from 2000 to \(2009,\) where \(x\) is the number of years after 2000 and \(y\) is pounds of cheese (in billions). Round answers to the nearest hundredth of a billion. (Source: National Agricultural Statistics Service, USDA) a. Approximate the number of pounds of cheese produced in the United States in 2000. b. Approximate the number of pounds of cheese produced in the United States in 2005. c. Use this function to estimate the pounds of cheese produced in the United States in 2015. d. From parts \((a),(b),\) and \((c),\) determine whether the number of pounds of cheese produced in the United States is increasing at a steady rate. Explain why or why not.
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