1. ALGEBRA 2: MODULE 4 LESSON 1 Solving Quadratic Equations by Factoring

2. Forms of a Quadratic Equation  Standard Form of a Quadratic Equation ax² + bx + c = 0, a ≠ 0  Factored Form of a Quadratic Equation a(x + p)(x + q) = 0, a ≠ 0 Factoring means to write the terms in multiplication form (as a product).  Zero Product Property If ab = 0 then either a = 0 or b = 0 (or both). The expression must be set equal to zero to use this property. Zero Product Example: Quadratic in Factored Form (x – 6) (x + 8) = 0 x – 6 = 0 or x + 8 = 0 x = 6 or x = 8

3. Solve a Quadratic in Factored Form  Solve (x – 11)(x – 1) = 0. This equation is already in factored form. All we have to do is use the Zero Product Property and set each factor equal to zero.  x – 11 = 0 or x – 1 = 0  x = 11 or x = 1  You may write your solutions as x = {1, 11}. Do not enclose the values of x in parentheses. They may be misconstrued as an ordered pair! But brackets may be used to signify a set of numbers.

4. Solve by Factoring 1. Write the quadratic equation in standard form: move all terms to one side of the equation, usually the left, using addition or subtraction such that one side of the equation is equal to zero. Simplify if needed. *To make the process easier, lets always try to make our quadratic term positive! [ax² + bx + c = 0] 2. Write the quadratic equation in factored form: factor the equation completely. [a(x + p)(x + q) = 0] 3. Set each factor equal to zero, and solve each equation. [If ab = 0 then either a = 0 or b = 0]

5. Solve Quadratics by Factoring- Example 1  Solve x 2 - 4x + 3 = 0.  The quadratic equation is in standard form (set equal to zero). This needs to be factored first.  Factor: x 2 - 4x + 3 = (x - 1)(x - 3) *You may want to use the X factors for this.  (x - 1)(x - 3) = 0  Set each factor equal to zero:  x - 1 = 0 or x - 3 = 0  Solve: x = 1 or x = 3, can also write {1, 3}

6. Solve Quadratics by Factoring- Example 2  Solve x 2 – 8 = 2x.  Write the equation in standard form (set the equation equal to zero). x 2 – 2x – 8 = 0  Factor: x 2 – 2x – 8 = (x – 4)(x + 2)  (x – 4)(x + 2) = 0  Set each factor equal to zero: x – 4 = 0 or x + 2 = 0  Solve: x = 4 or x = -2 or {4, -2}

7. Solve Quadratics by Factoring- Example 3  Solve x² + 3x = 0  Since it is written in standard form, factor it using GCF: x(x + 3) = 0.  Set all the factors equal to zero, and solve: x(x + 3) = 0 x = 0 or x + 3 = 0 x = 0 or x = –3 OR {0, -3}  Common mistakes on this type of problem:  Do not use X factors- that is for Trinomials.  Do not try to "solve" the equation for "x(x + 3) = 0" by dividing by x. Due to the fact that the zero product property says either factor could be zero or both dividing the x assumes that x is not zero. We don’t know that, and you lose one of your solutions. Therefore, we can't divide by zero…it against math laws!  Even though you are used to factors having variables and numbers (like the factor, x + 3), a factor might contain only a variable, such as x. Be sure to set this equal to zero also. If it said 5(x-2)(x+4) = 0, 5 cannot = 0!

8. Solve Quadratics by Factoring- Example 4  Solve (x + 2)(x + 3) = 20.  It is very common to see this type of problem, and say: It's already factored! So I'll set the factors equal to 0 and solve. Remember, the expression must be set equal to zero to use this property.  Simplify by the distributive property (x + 2)(x + 3) = 20  (x + 2)(x + 3) -20 = 0 x 2 + 5x + 6 - 20 = 0 x 2 + 5x – 14 = 0 It is in standard form, ready to factor (x + 7)(x – 2) = 0 Factored, can use X factors x + 7 = 0 or x – 2 = 0 Set each factor = to 0 and solve x = –7 or x = 2

9. Solve Quadratics by Factoring- Example 5  Solve 8x 2 = 26x - 15.  Write the equation in standard form (set the equation equal to zero). 8x 2 - 26x + 15 = 0  Factor: (4x – 3)(2x - 5) = 0  Set each factor equal to zero: 4x – 3 = 0 or 2x - 5 = 0  Solve: 4x – 3 = 0 or 2x - 5 = 0 4x = 3 or 2x = 5 x = ¾ or x = 5/2 -20 -6 2x2x- 5 4x4x8x28x2 -20x - 3-6x15

10. Back to the Garden Q: Suppose you are trying to create a garden. The length of the garden needs to be six feet longer than the width. You will be given 40 square feet of space. A: We know that the length is (x + 6) and the width is (x). The area is 40 sq. ft. If area equals length times width, then (x)(x+6) = 40. We want to find the values of that make this equation true. We can use the method of factoring to solve this problem. Simplify using the distributive property: x² + 6x = 40 Write it in standard form:x² + 6x – 40 = 0 Factor:(x + 10) (x – 4) = 0 Set each factor equal to zero:x + 10 = 0 or x – 4 = 0 Solve: x = -10 or x = 4 Since length cannot be negative, -10 cannot be a solution and the only solution is width = 4. Use 4 to find the length 4 + 6 = 10. The dimensions will be 4 feet x 10 feet. Image courtesy of Tom Curtis @ FreeDigitalPhotos.net