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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.

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Chapter 6 Factoring Polynomials

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  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:

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