5.2 Solving Quadratic Equations by Factoring

5.2 Solving Quadratic Equations by Factoring

4.3 Solving Quadratic Equations by Factoring Recall multiplying these binomials to get the standard form for the equation of a quadratic function: (x + 3)(x + 5) = x 2 + 5x +3x +15 2 x 8 x 15 The reverse of this process is called factoring. Writing a trinomial as a product of two binomials is called factoring. 2

x 8 x 15 (x + 3)(x + 5) Factor 2 x 12 x 28 Since the lead coefficient is 1, we need two numbers that multiply to 28 and add to 12. Factors of -28 -1,28 1,-28 -2,14 2,-14 -4,7 4,-7 Sum of Factors -3 Therefore:

27 2 -27 12 -12 3 x 12 x 28 ( x 2)( x 14) Factor the expression: = (x-3)(x+7) Cannot be factored Factoring a Trinomial when the lead coefficient is not 1. Factor:

2 3 x 17 x 10 We need a combination of factors of 3 and 10 that will give a middle term of 17. Our approach will guess and check. Here are some possible factorizations: (3x 10)( x 1) 3 x 2 13 x 10 (3x 5)( x 2) 3x 2 11x 10 (3 x 1)( x 10) 3 x 2 31x 10 (3x 2)( x 5) 3 x 2 17 x 10 This is the factorization we seek. Special Factoring Patterns you should remember: Pattern Name Difference of Two Squares

Perfect Square Trinomial Pattern 2 Example 2 x 9 ( x 3)( x 3) 2 a b (a b)(a b) 2 2 2

x 2 12 x 36 ( x 6) 2 2 2 2 x 2 8 x 16 ( x 4) 2 a 2ab b (a b) a 2ab b (a b) Factor the quadratic expression: 2 2 4 x 25 (2 x) 5

2 (2 x 5)(2 x 5) 2 2 9 y 24 y 16 (3 y ) 2(3 y )(4) 4 (3 y 4) 2 2 2 2 49r 14r 1 (7r ) 2(7 r )(1) 1 (7 r 1) 2

2 A monomial is an expression that has only one term. As a first step to factoring, you should check to see whether the terms have a common monomial factor. Factor: 2 5( x 4) 5 x 20 5( x 2)( x 2) 2 2u 2 8u 2u (u 4) 2 2

4 x 4 x 4 4( x x 1) You can use factoring to solve certain quadratic equation. A quadratic equation in one variable can be written in the form ax 2 bx c 0 where a 0 2 This is called the standard form of the equation: ax bx c 0 If this equation can be factored then we can use this zero product property. Zero Product Property Let A and B be real number or algebraic expressions. If AB = 0 the either A=0 or B=0 Solve:

2 x 3x 18 0 ( x 6)( x 3) 0 So, either (x+6)=0 Or (x 3)=0 The solutions are 6 and 3. These solutions are also called zeros of the function y x 2 3x 18 Notice the zeros are the x-intercepts of the graph of the function. x = -6 x=3

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