Intermediate Algebra – Student Workbook – Second Edition 2013

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$Lesson 2a – Assessment - Scottsdale Community College - Math Blog$

Lesson 5b - Solving Quadratic Equations Mini-Lesson Solving Quadratic Equations by Factoring So far, we have only used our graphing calculators to solve quadratic equations utilizing the Intersection process. There are other methods to solve quadratic equations. The first method we will discuss is the method of FACTORING. Before we jump into this process, you need to have some concept of what it means to FACTOR using numbers that are more familiar. Factoring Whole Numbers To FACTOR the number 60, you could write down a variety of responses some of which are below: 60 = 1 * 60 (not very interesting but true) 60 = 2 * 30 60 = 3 * 20 60 = 4 * 3 * 5 All of these are called FACTORIZATIONS of 60, meaning to write 60 as a product of some of the numbers that divide it evenly. The most basic factorization of 60 is as a product of its prime factors (remember that prime numbers are only divisible by themselves and 1). The PRIME FACTORIZATION of 60 is: 60 = 2 * 2 * 3 * 5 There is only one PRIME FACTORIZATION of 60 so we can now say that 60 is COMPLETELY FACTORED when we write it as 60 = 2 * 2 * 3 * 5. When we factor polynomial expressions, we use a similar process. For example, to factor the expression 24x 2 , we would first find the prime factorization of 24 and then factor x 2 . 24 = 2 * 2 * 2 * 3 and x 2 = x * x Putting these factorizations together, we obtain the following: 24x 2 = 2 * 2 * 2 * 3 * x * x Let’s see how the information above helps us to factor more complicated polynomial expressions and ultimately leads us to a second solution method for quadratic equations. Scottsdale Community College Page 212 Intermediate Algebra
Lesson 5b - Solving Quadratic Equations Mini-Lesson Factoring Quadratic Expressions and Solving Quadratic Equations by Factoring (Factoring using Greatest Common Factor Method) Let’s use the information on the previous page to help us FACTOR 3x 2 + 6x. The building blocks of 3x 2 + 6x are the terms 3x 2 and 6x. Each is written in FACTORED FORM below. 3x 2 = 3 * x * x and 6x = 3 * 2 * x Let’s rearrange these factorizations just slightly as follows: × 3x 2 = (3 * x) * x and 6x = (3 * x) * 2 We can see that (3 x) = 3x is a common FACTOR to both terms. In fact, 3x is the GREATEST COMMON FACTOR to both terms. Let’s rewrite the full expression with the terms in factored form and see how that helps us factor the expression: 3x 2 + 6x = (3 * x) * x + (3 * x) * 2 = (3x) * x + (3x) * 2 = (3x)(x + 2) Always check your factorization by multiplying the final result. (3x)(x + 2)=3x 2 + 6x CHECKS 3x 2 + 6x = (3x)(x + 2) is in COMPLETELY FACTORED FORM How can we use this factored form to solve quadratic equations such as 3x 2 + 6x = 0? To solve 3x 2 + 6x = 0, FACTOR the left side: (3x)(x + 2) = 0 Set each linear factor to 0: 3x = 0 OR x + 2 = 0 Solve each linear equation to get x = 0 OR x = -2 The solutions to 3x 2 + 6x = 0 are x = 0 or x = -2. If you graph the parabola 3x 2 + 6x, I see that it crosses the x-axis at (0, 0) and (-2, 0) which confirms the solutions found above. Scottsdale Community College Page 213 Intermediate Algebra
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Scottsdale Community College Interm
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Table of Contents Lesson 1a - Intro
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Lesson 1a - Introduction to Functio
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Lesson 1b - Linear Equations In the
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Lesson 1b - Linear Equations Mini-L
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Lesson 4a - Introduction to Logarit
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Lesson 6b - Rational Exponents & Ra
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Appendix A: You Try Answers Lesson
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Appendix A: You Try Answers You Try
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Intermediate Algebra – Student Workbook – Second Edition 2013

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