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World 1-8 Solving by Factoring Recall: it requires one equation to solve for one unknown. Traditionally we find solutions to equations by isolating x. 5x + 10 = 25 -10 -10 5x = 15 5 5 x=3 y2 + 25 = 61 -25 -25 y2 = √36 y=±6 x2 + x = 20 Cannot get x alone, by old methods. In the past solving an equation with an x2 and an x that didn’t cancel would not have been possible. But now you CAN by factoring. Consider the equation ab= 0 For this to be true. Either a=0, or b =0 ab= 0 ab=0 So a=0 and/or b=0 is (0)(b)=0 the solution. (a)(0)=0 works works as well 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 Solve by factoring 1. 3x – 6 = 0 3(x – 2)= 0 3=0 nonsense x–2=0 x=2 3. 4x + 20 = 0 4(x +5)= 0 x+5=0 x = -5 2. 7x – 7 = 0 7(x – 1)= 0 x–1=0 x=1 4. 16 – 12x = 0 4(4 – 3x)= 0 4 – 3x = 0 -4 -4 - 3x = -4 x = 4/3 Solve by factoring 1. 3x2 – 9x = 0 3x(x – 3)= 0 x= 0 x–3=0 x=3 or Two solutions 2. 4x – 3 = x + 3 -3 -3 4x – 6 = x -x -x 3x – 6 = 0 3(x – 2) = 0 x–2=0 x=2 Factor and Solve 1. x(x+1) – 5(x+1) = 0 (x + 1)(x- 5)= 0 x= -1 3. 2. x(x - 4) + 5(x - 4) Not an equation! or x = 5 x2 - 16= 0 (x -4)(x + 4)= 0 4. x2 – 5x + 14= 0 (x - 7 )(x + 2 )= 0 x = 7 or x = - 2 x= 4 or x = - 4 so x = ±4 Tougher One x3 + 2x2 -3x = 0 x(x2 +2x -3)= 0 x(x - 1)(x + 3)= 0 x= 0, x=1, and x= -3 are solutions Homework World 1-8