Lesson 7.4 Solving polynomial equations in factored form

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Presentation transcript:

Lesson 7.4 Solving polynomial equations in factored form EQ: How do you solve an equation in factored form? Objective: To use zero product property to solve factored equations Zero Product property: If a • b = 0, then either a = 0 or b = 0. Examples: What value of x would give a product of zero. 1. (x – 2)5 = 0 x = 2 x = –3 2. (x + 3)7 = 0 3. (7 – x)2 = 0 x = 7 x = 2 4. (3x – 6)5 = 0

Examples: Determine the values for x that makes each equation true. Either x + 3 = 0 or x – 2 = 0 1. (x + 3) (x – 2) = 0 –3 –3 +2 +2 x = –3 x = 2 Check –3 and 2 (–3 + 3)(–3 – 2) = 0 (0)(–5) = 0 (2 + 3) (2 – 2) = 0 Solutions –3 and 2 (5)(0) = 0 The graph of y = (x + 3)(x – 2) would be a parabola that crosses the x-axis at (–3,0) and (2,0) Two roots/solutions

Ex: Solve for x 2x + 4 = 0 or x – 3 = 0 (2x + 4)(x – 3) = 0 – 4 – 4 +3 +3 x = 3 2x = –4 2 2 x = –2 Solutions: –2 and 3 EX: is equal to (x + 5)(x + 5) = 0 (x + 5) is repeated. There will be only one answer. x + 5 = 0 –5 –5 The parabola touches the x-axis once at the point (–5,0) x = –5

Factor and solve the equation. Factoring is the opposite of the distributive property where the expression is factored. Factor 4x(x + 3) = 0 Zero Product prop 4x = 0 or x + 3 = 0 Roots are 0 and -3 x = 0 or x = -3 The solutions to polynomial equations are called roots.

Factor -2k(5k + 4) = 0 Zero product prop -2k = 0 or 5k + 4 = 0