2. A technique for factoring trinomials Lutheran High Westland Mathematics Department by way of Martin Luther High School Mathematics Department by way of A Catholic Nun

3.  Used to factor quadratics in the form of ax 2 + bx + c  For example: 3x 2 – 4x + 5 a = 3 b = -4 c = 5  This technique helps minimize the use of “guess and check”  You’re going to love this!

4.  Ten sample problems will be presented with explanation of the steps.  Don’t be intimidated. Practice makes this technique easy to remember!

5. 1.Find the Master Product by multiplying ‘a’ by ‘c’ MP 12 + 4 + 3 2.Find a pair of factors for the MP whose sum is ‘b’ 3.Find a pair of factors for ‘a’ where each will factor one of the factors found in step 2. 4.Write these factors with the variable in front of and opposite the numbers they factor from step 2. 5.Deflate: ‘4’ is replaced with ‘2’ as 4  2 is 2 ‘3’ is “replaced” with ‘3’ as 3  1 is 3 6.Write your answer: (x + 2)(2x + 3) x 2x / 2 / 3 1212

6. 1.Find the Master Product by multiplying ‘a’ by ‘c’ MP 30 - 15 - 2 2.Find a pair of factors for the MP whose sum is ‘b’ 3.Find a pair of factors for ‘a’ where each will factor one of the factors found in step 2. 4.Write these factors with the variable in front of and opposite the numbers they factor from step 2. 5.Deflate: ‘15’ is replaced with ‘5’ as 15  3 is 5 ‘2’ is “replaced” with ‘2’ as 2  1 is 2 6.Write your answer: (x - 5)(3x - 2) x 3x / 5 / 2 1313

7. 1.Find the Master Product by multiplying ‘a’ by ‘c’ MP -40 + 8 - 5 2.Find a pair of factors for the MP whose sum is ‘b’ 3.Find a pair of factors for ‘a’ where each will factor one of the factors found in step 2. 4.Write these factors with the variable in front of and opposite the numbers they factor from step 2. 5.Deflate: ‘8’ is replaced with ‘4’ as 8  2 is 4 ‘5’ is “replaced” with ‘5’ as 5  1 is 5 6.Write your answer: (x + 4)(2x - 5) x 2x / 4 / 5 1212

8. 1.Find the Master Product by multiplying ‘a’ by ‘c’ MP -24 - 24 + 1 2.Find a pair of factors for the MP whose sum is ‘b’ 3.Find a pair of factors for ‘a’ where each will factor one of the factors found in step 2. 4.Write these factors with the variable in front of and opposite the numbers they factor from step 2. 5.Deflate: ‘24’ is replaced with ‘8’ as 24  3 is 8 ‘1’ is “replaced” with ‘1’ as 1  1 is 1 6.Write your answer: (x - 8)(3x + 1) x 3x / 8 / 1 1313

9. 1.Find the Master Product by multiplying ‘a’ by ‘c’ MP 10 - 5 - 2 2.Find a pair of factors for the MP whose sum is ‘b’ 3.Find a pair of factors for ‘a’ where each will factor one of the factors found in step 2. 4.Write these factors with the variable in front of and opposite the numbers they factor from step 2. 5.Deflate: ‘5’ is replaced with ‘5’ as 5  1 is 5 ‘2’ is “replaced” with ‘2’ as 2  1 is 2 6.Write your answer: (x - 5)(x - 2) x x / 5 / 2 1111

10. 1.Find the Master Product by multiplying ‘a’ by ‘c’ MP -21 - 7 + 3 2.Find a pair of factors for the MP whose sum is ‘b’ 3.Find a pair of factors for ‘a’ where each will factor one of the factors found in step 2. 4.Write these factors with the variable in front of and opposite the numbers they factor from step 2. 5.Deflate: ‘7’ is replaced with ‘7’ as 7  1 is 7 ‘3’ is “replaced” with ‘3’ as 3  1 is 3 6.Write your answer: (x - 7)(x + 3) x x / 7 / 3 1111

11. 1.Find the Master Product by multiplying ‘a’ by ‘c’ MP -126 - 14 + 9 2.Find a pair of factors for the MP whose sum is ‘b’ 3.Find a pair of factors for ‘a’ where each will factor one of the factors found in step 2. 4.Write these factors with the variable in front of and opposite the numbers they factor from step 2. 5.Deflate: ‘14’ is replaced with ‘7’ as 14  2 is 7 ‘9’ is replaced with ‘3’ as 9  3 is 3 6.Write your answer: (3x - 7)(2x + 3) 3x 2x / 7 / 3 3232

12. 1.Find the Master Product by multiplying ‘a’ by ‘c’ MP -20 +10 - 2 2.Find a pair of factors for the MP whose sum is ‘b’ 3.Find a pair of factors for ‘a’ where each will factor one of the factors found in step 2. 4.Write these factors with the variable in front of and opposite the numbers they factor from step 2. 5.Deflate: ‘10’ is replaced with ‘5’ as 10  2 is 5 ‘-2’ is replaced with ‘1’ as -2  -2 is 1 6.Write your answer: (-2x + 5)(2x + 1) -2x 2x / 5 / +1 -2 2

13. 1.Find the Master Product by multiplying ‘a’ by ‘c’ MP 100 + 10 + 10 2.Find a pair of factors for the MP whose sum is ‘b’ 3.Find a pair of factors for ‘a’ where each will factor one of the factors found in step 2. 4.Write these factors with the variable in front of and opposite the numbers they factor from step 2. 5.Deflate: ‘10’ is replaced with ‘5’ as 10  2 is 5 ‘10’ is replaced with ‘5’ as 10  2 is 5 6.Write your answer: (2x + 5)(2x + 5) 2x 2x / 5 / 5 2222

14. 1.Find the Master Product by multiplying ‘a’ by ‘c’ MP -100 + 10 - 10 2.Find a pair of factors for the MP whose sum is ‘b’ 3.Find a pair of factors for ‘a’ where each will factor one of the factors found in step 2. 4.Write these factors with the variable in front of and opposite the numbers they factor from step 2. 5.Deflate: ‘10’ is replaced with ‘5’ as 10  2 is 5 ‘10’ is replaced with ‘5’ as 10  2 is 5 6.Write your answer: (2x + 5)(2x - 5) 2x 2x / 5 / 5 2222