1 / 9

Chapter 6 Factoring Polynomials

Chapter 6 Factoring Polynomials. Section 3 Factor Trinomials of the Form ax 2 + bx + c , a ≠1. Section 6.3 Objectives. 1 Factor ax 2 + bx + c , a ≠1, Using Trial and Error 2 Factor ax 2 + bx + c , a ≠1, Using Grouping. ( __ x + )( __ x + ) = ax 2 + bx + c.

Download Presentation

Chapter 6 Factoring Polynomials

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 6 Factoring Polynomials Section 3 Factor Trinomials of the Form ax2 + bx + c,a ≠1

  2. Section 6.3 Objectives 1 Factor ax2 + bx + c, a ≠1, Using Trial and Error 2 Factor ax2 + bx + c, a ≠1, Using Grouping

  3. ( __x + )( __x + ) = ax2 + bx + c Trinomials of the Form ax2 + bx+c, a  1 When factoring trinomials of the form ax2 + bx + c where a is not equal to 1, there are two methods that can be used: 1. Trial and error; 2. Factoring by grouping Factoring ax2 + bx + c, a 1 Using Trial and Error: a, b, and c Have No Common Factors Step 1: List the possibilities for the first terms of each binomial whose product is ax2. Step 2: List the possibilities for the last terms of each binomial whose product is c. Step 3: Write out all the combinations of factors found in Steps 1 and 2. Multiply the binomials out until a product is found that equals the trinomial. ( __x + )( __x + ) = ax2 + bx + c

  4. Correct The Trial and Error Method Example: Factor by trial and error: 3x2+4x +1 The Factors of 1 are: The Factors of 3 are: 1 and 3 1 and 1 Possible Factors: Middle Term: (x + 1)(3x+ 1) 4x (x + 1)(3x + 1) = 3x2 + 4x + 1 Check: 

  5. Correct The Trial and Error Method Example: Factor: 5x2 + 2x  7 The sign of one factor will be positive, the sign of the other factor will be negative. The Factors of 7 Are: The Factors of 5 Are: 1 and 5 1 and 7 Possible Factors: Middle Term: (x + 1)(5x 7) – 2x 2x (x 1)(5x + 7) 5x2 + 2x  7 = (x – 1)(5x + 7) (x – 1)(5x + 7) = 5x2 + 2x  7 Check: 

  6. Correct The Trial and Error Method Example: Factor by trial and error: 9x2 13x + 4 The Factors of 4 are: The Factors of 9 are: 1 and 9 1 and 4 3 and 3 2 and 2 Possible Factors: Middle Term: (x – 1)(9x – 4) – 13x – 20x (x – 2)(9x – 2) – 15x (3x – 1)(3x – 4) – 12x (3x – 2)(3x – 2) 9x2 13x + 4 = (x – 1)(9x – 4) Check: 

  7. Factoring by Grouping Factoring ax2 + bx + c, a 1 by Grouping: a, b, and c Have No Common Factors Step 1: Find the value of ac. Step 2: Find the pair of integers, m and n, whose product is ac and whose sum is b. Step 3: Write ax2+ bx + c = ax2 + mx + nx + c. Step 4: Factor the expression in Step 3 by grouping. Step 5: Check bymultiplying the factors.

  8. Factoring by Grouping Example: Factor by grouping: 2x2 + 9x + 4 The value of ac = 8. 1 8 Find two numbers that we can multiply together to get 8 and add together to get 9. ___ × ___ = 8 1 8 ___ + ___ = 9 2x2 + 9x + 4 = 2x2 + x + 8x + 4 Rewrite 9x as x + 8x. = x(2x + 1) + 4(2x + 1) Factor by grouping. = (2x + 1)(x + 4) Factor out (2x + 1). (2x + 1)(x + 4) = 2x2 + 8x + 1x + 4 Check: = 2x2 + 9x + 4 

  9. Factoring by Grouping Example: Factor by grouping: 8x2 + 10x 3 The value of ac is  24. ( 2) 12 ____ × ____ =  24 Find two numbers that we can multiply together to get  24 and add together to get 10. ( 2) 12 ____ + ____ = 10 8x2 + 10x 3 = 8x2 + 12x 2x  3 Rewrite 10x as 12x + (2x). = 4x(2x + 3)  1(2x + 3) Factor by grouping. = (2x + 3)(4x 1) Factor out (2x + 3). (2x + 3)(4x 1) = 8x2 + 10x 3  Check:

More Related