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12-5 Factoring ax ² + bx + c

12-5 Factoring ax ² + bx + c. When you multiplied, the _ was ______________ the _ and became _. It was then ____________ . So now we have to find the correct ___________ for the ____ and ________. The only way. 4. Multiply. multiplied times. 2. ( 2n + 3 )( n + 4 ). 8.

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12-5 Factoring ax ² + bx + c

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  1. 12-5 Factoring ax² + bx + c When you multiplied, the _ was ______________ the _ and became _. It was then ____________ . So now we have to find the correct ___________ for the ____ and ________. The only way 4 Multiply multiplied times 2 (2n+ 3)(n+ 4) 8 added to the3 2n²+ 8n + 3n+ 12 2n²+ 11n+ 12 combination 1st last term Factor ( n+ )( n+ ) 2&1 to factor 2 is _____ but this must be ________ by the __________ , then ___ ___________ together to get the ____________. add the products multiplied factors of 12 middle term

  2. 12-5 Factoring ax² + bx + c 2n² + 11n+ 12 combination for 1st term combination for last term Factor 2 · 1 · 6 4 3 = 4 = (?n+?)(?n+?) 10 Not the middle term O other bin. 4 = 3 = 2 · 1 · 8 3 I one bin. 11 ( n+ )( n+ ) 2 3 1 4

  3. Factor 1st term Combo. Last term Combo. 1. 2n²- 3n- 20 4 = -5 = 8 -5 2 · 1 · ( n+ )( n+ ) 2 5 1 -4 3 No, wrong sign Outside -4 = 5 = other bin. -8 5 2 · 1 · Inside one bin. ( n+ )(n ) 2 5 - 4 -3 2n²- 8n + 5n- 20 2n²- 3n- 20

  4. 2. 6y²- 29y- 5 1st term Combo. Last term Combo. 1 = -5 = 2 -15 2 · 3 · 6y²+ y - 30y- 5 -13 No, # too large check 1 = -5 = 3 -10 3 · 2 · ( y - )( y + ) 1 5 6 1 -7 Even larger Outside other bin. 1 = -5 = 1 -30 1 · 6 · Inside one bin. -29

  5. 3. Is 2x²+ 5x+ 3 factorable? If it is, factor it, if not tell why. 1st term Last term 3 = 1 = 6 1 2 · 1 · ( x+ )( x + ) 2 3 1 1 7 no Outside other bin. 1 = 3 = 2 3 2 · 1 · Inside one bin. 5

  6. Solve each equation. 1st factor, Same as # 1 4. 2n² - 3n - 20 = 0 ( n +)(n ) = 0 2 5 - 4 Set each binomial = 0 n -4 = 0 2n + 5 = 0 -5 -5 4 4 2n = -5 n = 4 Solve both equations. 2 2 n = -2.5

  7. 5. 6y² - 29y -5 = 0 1st factor, which we already did in problem 2. (y -)(y + ) = 0 5 6 1 y - 5= 0 6y + 1= 0 Set each binomial = 0 6y = -1 y = 5 _1 6 y = Solve both equations. 1) GCF 6. 12y³ - 58y² - 10y = 0 2y (6y² - 29y -5) = 0 2) 2 binomials 2y (y - 5)(6y + 1) = 0 3) Solve _ 1 6 y = 0, 5, 2y=0 (y - 5)=0 (6y + 1)=0

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