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6.3 Factoring Trinomials II

Ax 2 + bx + c. 6.3 Factoring Trinomials II. Factoring Trinomials Review. X 2 + 6x + 5 (x )(x ) Find factors of 5 that add to 6: 1*6 = 6 1+6 = 7 2*3 = 6 2+3 = 5 (x + 2)(x + 3). Factoring Trinomials where a ≠ 1. Follow these steps:

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6.3 Factoring Trinomials II

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  1. Ax2 + bx + c 6.3 Factoring Trinomials II

  2. Factoring Trinomials Review X2 + 6x + 5 (x )(x ) Find factors of 5 that add to 6: 1*6 = 6 1+6 = 7 2*3 = 6 2+3 = 5 (x + 2)(x + 3)

  3. Factoring Trinomials where a ≠ 1 Follow these steps: 1. Find two numbers that multiply to ac and add to b for ax2 + bx + c 2. Replace bx with the sum of the 2 factors found in step 1. ie: ax2 + bx + c becomes ax2 + mx + nx + c, where m and n are the factors found in step 1. 3. Use grouping to factor this expression into 2 binomials

  4. 2x2 + 5x + 2 Step 1: ac = 2*2 = 4 1*4 = 4 1+4 = 5 2*2 = 4 2+2 = 4 m = 1 and n = 4 Step 2: Rewrite our trinomial by expanding bx 2x2 + 1x + 4x + 2 Step 3: Group and Factor (2x2 + 1x) + (4x + 2) x(2x + 1) + 2( 2x + 1) (2x + 1) (x + 2)

  5. 2x2 + 5x + 2 Questions for thought: Does it matter which order the new factors are entered into the polynomial? Do the parenthesis still need to be the same? Will signs continue to matter when finding m and n? Does it matter how we group the terms for factoring?

  6. 3z2 + z – 2 Step 1: ac = 3*-2 = -6 -1*6 = -6 -1+6 = 5 1* -6 = -6 1+-6 = -5-2*3 = -6 -2+3 = 1 2* -3 = -6 2+-3 = -1 m = -2 and n = 3 Step 2: Rewrite our trinomial by expanding bx 3z2 + 3z – 2z – 2 Step 3: Group and Factor (3z2 + 3z) + (-2z - 2) 3z(z + 1) - 2( z + 1) (z + 1) (3z - 2)

  7. 3z2 + z – 2 Step 1: ac = 3*2 = 6 -1*6 = -6 -1+7 = 6 1* -6 = -6 1+-7 = -6-2*3 = -6 -2+3 = 1 2* -3 = -6 2+-3 = -1 m = -2 and n = 3 Step 2: Rewrite our trinomial by expanding bx 3z2+ 3z – 2z – 2 Notice that I changed the order of m and n between step 1 and step 2. Why do you think I did this? Do you have to change the order to get the correct answer?

  8. 3z2 + z – 2 What are the 3 steps for solving this quadratic equation? Step 1: Multiply a*c. Find the factors that multiply to ac and add to b Step 2: Expand bx to equal mx + nx Step 3: Group and Factor

  9. 4x3 – 22x2 + 30x Step 0: Factor out the GCF: 2x 2x(2x2 – 11x + 15) Step 1: a*c = 30 -1*-30 = 30 -1+-30 = -31 -2*-15 = 30 -2+-15 = -17 -3*-10 = 30 -3+-10 = -13 -5*-6 = 30 -5+-6 = -11

  10. 4x3 – 22x2 + 30x Step 0: Factor out the GCF: 2x 2x(2x2 – 11x + 15) Step 1: a*c = 30 -1*-30 = 30 -1+-30 = -31 -2*-15 = 30 -2+-15 = -17 -3*-10 = 30 -3+-10 = -13 -5*-6 = 30 -5+-6 = -11

  11. 4x3 – 22x2 + 30x Step 2: Expand bx to equal mx + nx -11x = -5x + -6x 2x(2x2 – 5x – 6x + 15) Step 3: Group and Factor 2x((2x2 – 5x )(– 6x + 15)) 2x(x(2x – 5) -3(2x – 5)) 2x(2x – 5) (x – 3)

  12. 4x3 – 22x2 + 30x Step 2: Expand bx to equal mx + nx -11x = -5x + -6x 2x(2x2 – 5x – 6x + 15) Step 3: Group and Factor 2x((2x2 – 5x )(– 6x + 15)) 2x(x(2x – 5) -3(2x – 5)) Note: The Parenthesis are the Same 2x(2x – 5) (x – 3)

  13. Practice 1. 3x2 + 5x + 2 2. 6x2 + 7x – 3 3. 6 + 4y2 – 11y

  14. Practice 1. 3x2 + 5x + 2 (3x + 2)(x + 1) 2. 6x2 + 7x – 3 (3x – 1)(2x + 3) 3. 6 + 4y2 – 11y (4y – 3)(y – 2)

  15. Review What is Step 0? When do you need to include this step? When will your factors both be negative? When will you have one negative and one positive factor? How do you check your answers?

  16. ??? Questions ???

  17. 6.4 Special Types of Factoring

  18. 1. Differnce of Squares 2. Perfect Square Trinomials (Sum and Difference of Cubes is not included)

  19. Difference of Squares Think back to Chapter 5. What happened when we multiplied a sum and difference? (a – b)(a + b) = a2 – b2 So, the reverse is also true. a2 – b2 = (a – b)(a + b)

  20. x2 – 25 Notice that we do not have a bx term. This means that we only have the F and L in foil; therefore, none of the procedures from 6.1, 6.2, or 6.3 will work. We need to use a2 – b2 = (a – b)(a + b) where a = x and b = 5 X2 – 25 = (x – 5)(x + 5)

  21. x2 – 36 We need to use a2 – b2 = (a – b)(a + b) where a = x and b = 6 X2 – 36 = (x – 6)(x + 6)

  22. Practice 4x2 – 9 100 – 16t2 49y2 – 64z2

  23. Practice 4x2 – 9 a = 2x, b = 3 (2x – 3) (2x + 3) 100 – 16t2 a = 10, b = 4t (10 – 4t) (10 + 4t) 49y2 – 64z2 a = 7y, b = 8z (7y – 8z) (7y + 8z)

  24. Perfect Square Trinomials Think back to Chapter 5. What happened when we squared a binomial? (a + b)2 = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2 So, the reverse is also true. a2 + 2ab + b2 = (a + b)2 a2 – 2ab + b2 = (a – b)2

  25. x2 + 10x + 25 This can be worked 2 different ways The first way is the simplest, but depends on whether you recognize the equation as a perfect square trinomial. a2 + 2ab + b2 = (a + b)2 Where a = x and b = 5 x2 + 10x + 25 = (x + 5)2

  26. x2 + 10x + 25 This can be worked 2 different ways The second way is to use the method we learned in 6.2 x2 + 10x + 25 5*5 = 25 and 5+5 = 10 (x + 5) (x + 5) or (x + 5)2

  27. 4x2 - 4x + 1 This can be worked 2 different ways The first way is the simplest, but depends on whether you recognize the equation as a perfect square trinomial. a2 + 2ab + b2 = (a + b)2 Where a = 2x and b = 1 4x2 - 4x + 1 = (2x – 1)2

  28. 4x2 - 4x + 1 This time we need to use the 6.3 method 4*1 = 4 -2 * -2 = 4 and -2 + -2 = -4 (4x2 – 2x) ( – 2x + 1) 2x(2x – 1) – 1(2x – 1) (2x – 1) (2x – 1) or (2x – 1)2

  29. Practice x2 – 4xy + 4y2 9a2 – 60a + 100 25y2 + 20yz + 4z2

  30. Practice x2 – 4xy + 4y2 a = x, b = 2y (x – 2y)2 9a2 – 60a + 100 a = 3a, b = 10 (3a – 10) 25y2 + 20yz + 4z2 a = 5y, b = 2z (5y + 2z)

  31. Review What methods can you use to factor a Difference of Squares? What methods can you use to factor a Perfect Square Trinomial? What clues should you look for to identify a Difference of Squares? What clues should you look for to identify a Perfect Square Trinomial?

  32. ??? Questions ???

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