Download presentation
Presentation is loading. Please wait.
Published byHollie Montgomery Modified over 8 years ago
1
Chapter 5 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 5-1 Factoring
2
2 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 5-2 5.1 – Factoring a Monomial from a Polynomial 5.2 – Factoring by Grouping 5.3 – Factoring Trinomials of the Form ax 2 + bx + c, a = 1 5.4 – Factoring Trinomials of the Form ax 2 + bx + c, a ≠ 1 5.5 – Special Factoring Formulas and a General Review of Factoring 5.6 – Solving Quadratic Equations Using Factoring 5.7 – Applications of Quadratic Equations Chapter Sections
3
3 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 5-3 Solving Quadratic Equations Using Factoring
4
4 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 5-4 Quadratic Equation A A quadratic equation is an equation that contains a second-degree term and no term of a higher degree. Examples: x 2 + 4x - 12 = 0 2x 2 – 5x = 0 Quadratic equations have the form ax 2 + bx + c = 0 where a, b, and c are real numbers, a 0
5
5 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 5-5 Zero-Factor Property To solve a quadratic equation by factoring, the zero-factor property is used. Example: Solve the equation (x + 3)(x + 4) = 0 x + 3 = 0 or x+ 4 = 0 x + 3 – 3 = 0 - 3 x + 4 – 4 = 0 - 4 x = - 3 x = -4 If m · n = 0, then m = 0 or n = 0
6
6 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 5-6 Solving with Factoring 1.Write the equation in standard form with the squared term having a positive coefficient. This will result in one side of the equation being 0. 2.Factor the side of the equation that is not 0. 3.Set each factor containing a variable equal to 0 and solve each equation. 4.Check each solution found in step 3 in the original equation.
7
7 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 5-7 Solving with Factoring Example: Solve the following equation. a.) 3x 2 = 12x 3x 2 - 12x = 12x - 12x 3x 2 - 12x = 0 3x(x – 4) = 0 Now set each factor equal to 0. 3x = 0 or x – 4 = 0 x = 0 x = 4 The solutions are 0 and 4. Check by substituting x = 0, then x = 4, into 3x 2 = 12x
8
8 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 5-8 Solving with Factoring b.)x 2 + 10x + 28 = 4 x 2 + 10x + 24 = 0 (x + 4) (x + 6) = 0 x + 4 = 0 orx + 6 = 0 The solutions are – 4 and -6. x = - 4 x = – 6
Similar presentations
© 2024 SlidePlayer.com. Inc.
All rights reserved.