University:
High School
Course:
High School Mathematics
Academic year:

2022
Views:

390

Pages:

4
Author:

fwlsodtqx0

Solving Polynomial Equations in Factored Form Content: Factored form Roots: -Solution Repeated roots: -when the same solution appears two or more times Core Concepts Zero-Product Property Words: If the product of two real numbers is 0, then at least one of the numbers is 0. Algebra: If a and b are real numbers and a * b = 0, then a = 0 or b = 0. Identifying Solutions in a graph... will cross the x-axis Visual Elements: Two graphs of polynomial functions are shown. The x-intercepts of the graphs are marked with arrows and labeled with their corresponding x-values. 7.4 Extra Practice Instructions: In Exercises 1-12, solve the equation. Solutions: x(x+5) = 0 x = 0 (Solution) x + 5 = 0 x = -5 (Solution) a(a-12) = 0 a = 0 a - 12 = 0 a = 12 5p(p-2) = 0 5p = 0 p = 0 p - 2 = 0 p = 2 (c-2)(c+1) = 0 c - 2 = 0 c = 2 c + 1 = 0 c = -1 (2b-6)(3b+18) = 0 2b - 6 = 0 2b = 6 b = 3 3b + 18 = 0 3b = -18 b = -6 (3-5s)(-3+5s) = 0 3 - 5s = 0 -5s = -3 s = 3/5 -3 + 5s = 0 5s = 3 s = 3/5 (x-3)² = 0 (x-3)(x-3) = 0 x - 3 = 0 x = 3 (This is a double solution) (3d+7)(5d-6) = 0 3d + 7 = 0 3d = -7 d = -7/3 5d - 6 = 0 5d = 6 d = 6/5 (2t+8)(2t-8) = 0 2t + 8 = 0 2t = -8 t = -4 2t - 8 = 0 2t = 8 t = 4 (x+4)²(x+1) = 0 (x+4)(x+4)(x+1) = 0 x + 4 = 0 x = -4 (Double Solution) x + 1 = 0 x = -1 3 Solutions g(6-3g)(6+3g) = 0 g = 0 6 - 3g = 0 -3g = -6 g = 2 6 + 3g = 0 3g = -6 g = -2 (4-k)(8+3k)(3k-2) = 0 4 - k = 0 -k = -4 k = 4 8 + 3k = 0 3k = -8 k = -8/3 3k - 2 = 0 3k = 2 k = 2/3