1. 10.4 Solving Polynomial Equations in Factored Form Objective: I will use the zero-product property to find solutions to polynomial equations that are factored into the product of linear equations.

2. Vocabulary 1. Zero-product property - for all real numbers, if ab = 0, then a = 0 or b = 0 2. Repeated factor - polynomial with more than one of the same factor

3. Using the Zero-Product Property Solve the equation (x - 4)(x + 3) = 0 a = x - 4, b = x + 3 Using the zero-product property: x - 4 = 0orx + 3 = 0 +4 +4 - 3 - 3 x = 4 x = -3 The solutions are -3 and 4.

4. Guided Practice 1.Solve: (x - 8)(x + 4) = 0 2.Solve: (x - 14)(x - 23) = 0

5. Solving a Repeated-Factor Equation Solve (x - 4) 2 = 0 This equation has a repeated factor because (x - 4) 2 = (x - 4)(x - 4) So x - 4 = 0 +4 +4 x = 4 The solution is 4. Same factor

6. Guided Practice 1.Solve: (x + 5) 2 = 0 2.Solve: (x - 9) 3 = 0

7. Solving a Factored Cubic Equation Solve (4x + 3)(2x - 1)(x - 5) = 0 a polynomial equation has as many solutions as it does linear factors 4x + 3 = 02x - 1 = 0x - 5 = 0 - 3 - 3 + 1 + 1 + 5 + 5 4x = -3 2x = 1 x = 5 4 4 2 2 x = x = The solution is,, and 5.

8. Guided Practice 1.Solve: (2x + 4)(3x - 9)(x - 1) = 0 2.Solve: (5x - 30)(4x - 8)(x + 1) 2 = 0

9. Factoring a Polynomial, then Solving Factor and Solve: x 2 - 5x + 4 = 0 Factor: (x - 1)(x - 4) = 0 Solve:x - 1 = 0x - 4 = 0 +1 +1 +4 +4 x = 1 x = 4 Solution: x = 1, 4

10. Guided Practice 1.Factor and Solve: x 2 + 3x + 2 = 0 2.Factor and Solve: x 2 - 2x - 63 = 0

11. Independent Practice Solve. 1.(x - 3)(x + 4) = 0 2.(x + 8) 2 = 0 3.(5x + 10)(2x - 16)(x - 4) = 0 4.x 3 + 2x 2 + 2x + 4 = 0

12. Factors, Solutions, and x-intercepts For any quadratic polynomial ax 2 + bx + c = 0: If (x - p) is a factor of ax 2 + bx + c = 0 Then x = p is a solution of ax 2 + bx + c = 0 Then p is an x-intercept of the graph of ax 2 + bx + c = 0

13. Sketching Graphs of Quadratic Equations Using Factors and x-intercepts Sketch a graph of y = (x + 2)(x - 4) First find the x-intercepts x + 2 = 0x - 4 = 0 - 2 - 2 + 4 + 4 x = -2 x = 4 x-intercepts: (-2,0) and (4,0)

14. Sketching Graphs of Quadratic Equations Using Factors and x-intercepts Continued… Second find the vertex The x-value of the vertex is the average of the two x-intercepts x = 4 + (-2) = 2 = 1 2 2 Substitute x into the equation to find y y = (x + 2)(x - 4) = (1 + 2)(1 - 4) = (3)(-3) y = -9 Vertex: (1,-9)

15. Sketching Graphs of Quadratic Equations Using Factors and x-intercepts Continued… Sketch the graph xy

16. Using a Quadratic Model An arch is modeled by the equation y = -0.25(x - 6)(x + 6), with x and y measured in feet. How wide is the arch at the base? How high is the arch?

17. Using a Quadratic Model Solution x-intercepts (where y = 0): x - 6 = 0x + 6 = 0 + 6 + 6 - 6 - 6 x = 6 x = -6 -6, 6 Vertex (maximum height): x = 0 (average of x-values) y = -0.25(0 - 6)(0 + 6) = -0.25(-6)(6) = 9 Maximum height = 9 ft

18. Using a Quadratic Model Continued… Arch Width (horizontal distance where y = 0): x 2 - x 1 = 6 - (-6) = 6 + 6 = 12 Arch Width = 12 ft xy