Multiply (2n + 3)(n + 4) 2n² + 8n + 3n + 12 2n² + 11n + 12 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 the 3 2n² + 8n + 3n + 12 2n² + 11n + 12 combination 1st last term Factor to factor 2 is _____ but this must be ________ by the __________ , then ___ ___________ together to get the ____________. ( n+ )( n+ ) 2&1 add the products multiplied factors of 12 middle term

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

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

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

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

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

5. 6y² - 29y -5 = 0 (y - )( y + ) = 0 5 6 1 y - 5 = 0 6y + 1 = 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. 6. 12y³ - 58y² - 10y = 0 1) GCF 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