ࡱ> CEB{ bjbjVV 4r<<p FPP8SA""""dAfAfAfAfAfAfA$DFA"""""AUA "dRdA "dA R="@@-p>PAA0A>-G-G<@-G@@"" """""AAv&"""A""""-G"""""""""P Y: Algebra 2 Name ___________________________ Unit 11.10 Period 1 2 3 4 5 6 7 Factor and Solve a Trinomial Quadratic with a Leading Coefficient > 1 NOTES: Recall Multiply the binomials.  EMBED Equation.3  These are the factors of the trinomial answer your will get after FOILing the binomials. Steps to Factoring and Solving a Trinomial when the Leading Coefficient is > 1  EMBED Equation.3  Ex:  EMBED Equation.3  1. 2. 3. 4. 5. EXAMPLES: FACTOR SOLVE 1.  EMBED Equation.3 = 0 FACTOR SOLVE 2.  EMBED Equation.3  = 0 3.  EMBED Equation.3 = 0 4.  EMBED Equation.3 = 0 5.  EMBED Equation.3  6. 50x2 - 2 = 0 Algebra 2 Name_______________________ 11.10Factoring and Solving a Quadratic Period 1 2 3 4 5 6 7 Equation with a Leading Coefficient > 1 ASSIGNMENT Factor and solve the quadratic equation. C and B Level FACTOR SOLVE 1.  EMBED Equation.3 = 0  FACTOR SOLVE 2.  EMBED Equation.3 = 0  FACTOR SOLVE 3.  EMBED Equation.3 = 0  FACTOR SOLVE 4.  EMBED Equation.3 = 0  FACTOR SOLVE 5.  EMBED Equation.3 = 0  FACTOR SOLVE 6.  EMBED Equation.3 = 0  FACTOR SOLVE 7.  EMBED Equation.3 = 0  FACTOR SOLVE 8.  EMBED Equation.3 = 0  Factor out the GCF, factor the remaining trinomial, and then solve the quadratic equation. A Level FACTOR SOLVE 9.  EMBED Equation.3 = 0  FACTOR SOLVE 10.  EMBED Equation.3 = 0  FACTOR SOLVE 11.  EMBED Equation.3 = 0  FACTOR SOLVE 12.  EMBED Equation.3 = 0  FACTOR SOLVE 13.  EMBED Equation.3 = 0  FACTOR SOLVE 14.  EMBED Equation.3 = 0  FACTOR SOLVE 15.  EMBED Equation.3 = 0  11.10 Practice Test 1C. Factor.  EMBED Equation.3  1B. Solve by factoring.  EMBED Equation.3  1A. Solve by factoring.  EMBED Equation.3   RANDOM ANSWERS (x+1)(3x+4)=0 x = -1, -4/3 (2x-3)(3x+4)=0 x = 3/2, -4/3 (x-1)(5x-12)=0 x = 1, 12/5(x+1)(2x+3)=0 x = -1, -3/2 (x+7)(2x+1)=0 x = -7, -1/2 (x-1)(7x-2)=0 x = 1, 2/7 (7x-1)(7x+1)=0 x = -1/7, 1/74(2x-1)(2x+1)=0 x = -1/2,  (5x+3)(5x-3)=0 x = -3/5, 3/5(x+6)(2x+1)=0 x = -6, -1/2 (3x-2)(3x+2)=0 x = -2/3, 2/31C (x-4)(2x+3) 5(2x-3)(x+1)=0 x = -1, 3/23(2x-3)(3x+4)=0 x = 3/2, -4/31B (x+4)(5x+1)=0 x = -4, -1/5 (2x-9)(2x-1)=0 x = 9/2, 4(2x-1)(3x+1)=0 x = -1/3, 1A 2(x-3)(2x+7)=0 x = 3, -7/2     Reference Glencoe Textbook: Page 493-498 Chapter 8-4  6<lm úq`XOFOFOh >*OJQJh8x>*OJQJhD1OJQJ!jhD1hD1EHOJQJU%j+HM hD1CJOJPJQJUVhu1hD1OJQJjhu1hD1OJQJU%jh 5OJQJUmHnHuhD15OJQJh8x5OJQJh 5OJQJhv5OJQJh OJQJhvhvOJQJhnqOJQJhvOJQJh*fOJQJ2m    # $ ' ( + , / 0 : O P dhgdD1gd         / 0 : @ F I N ɸ簨{jbbYPBBh h 5>*OJQJh 5OJQJh8x5OJQJh8xOJQJ!jh8xh EHOJQJU%jAI h CJOJPJQJUVhu1h OJQJjhu1h OJQJUh OJQJhD1OJQJ!jh8xhD1EHOJQJU%jXAI hD1CJOJPJQJUVhu1hD1OJQJjhu1hD1OJQJUhD1>*OJQJN O P S T g h i j m o v w x ~ ʷ՞䞖l[՞!j h8xh8xEHOJQJU%j7BI h8xCJOJPJQJUVh h 5>*OJQJh 5OJQJh V.OJQJh OJQJ!j/h8xh8xEHOJQJU%jBI h8xCJOJPJQJUVhu1h8xOJQJjhu1h8xOJQJUh8xOJQJh8x5OJQJh h OJQJP p q r s t u v w x gd     ! 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