1. World 1-8 Solving by Factoring

2. y 2 = √36 5x = 15 Recall: it requires one equation to solve for one unknown. Traditionally we find solutions to equations by isolating x. Cannot get x alone, by old methods. 5x + 10 = 25 -10 5 5 x = 3 y 2 + 25 = 61 -25 y = ± 6 x 2 + x = 20 In the past solving an equation with an x 2 and an x that didn’t cancel would not have been possible. But now you CAN by factoring.

3. works Consider the equation ab= 0 For this to be true. Either a=0, or b =0 ab= 0 (0)(b)=0 ab=0 (a)(0)=0 works as well So a=0 and/or b=0 is the solution. STEPS to use this property 1. Rearrange the equation to let one side = 0 2. Factor the expression 3. If a factor has a variable let it = 0 4. Solve for the variable

4. Solve by factoring 1. 3x – 6 = 0 3(x – 2)= 0 3 = 0 nonsense x – 2 = 0 x = 2 eg. 1 2. 7x – 7 = 0 7(x – 1)= 0 x – 1 = 0 x = 1 3. 4x + 20 = 0 4(x +5)= 0 x + 5 = 0 x = -5 4. 4. 16 – 12x = 0 4(4 – 3x)= 0 4 – 3x = 0 -4 - 3x = -4 x = 4/3

5. Solve by factoring eg. 2 1. 3x 2 – 9x = 0 3x(x – 3)= 0 x= 0 or x – 3 = 0 x = 3 Two solutions 2. 4x – 3 = x + 3 - 3 4x – 6 = x - x 3x – 6 = 0 3(x – 2) = 0 x – 2 = 0 x = 2

6. (x - )(x + )= 0 Factor and Solve eg. 3 1. x(x+1) – 5(x+1) = 0 (x + 1)(x- 5)= 0 x= -1 orx = 5 2. x(x - 4) + 5(x - 4) Not an equation! 3. x 2 - 16= 0 (x -4)(x + 4)= 0 x= 4 orx = - 4 so x = ±4 4. 4. x 2 – 5x + 14= 0 7 2 x = 7 or x = - 2

7. Tougher One eg. 4 x 3 + 2x 2 -3x = 0 x(x 2 +2x -3)= 0 x(x - )(x + )= 01 3 x= 0, x=1, and x= -3 are solutions

8. Homework World 1-8