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Chapter 6 Review Factoring I (Factoring Polynomials)

Chapter 6 Review Factoring I (Factoring Polynomials). 6.1 Greatest Common Factor & Factoring by Grouping 6.2 Factoring Trinomials of the Form ax 2 + bx + c 6.3 Factor Trinomials of the Form ax 2 + bx + c , a ≠1. Section 6.1 Objectives.

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Chapter 6 Review Factoring I (Factoring Polynomials)

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  1. Chapter 6 Review Factoring I (Factoring Polynomials) • 6.1 Greatest Common Factor & Factoring by Grouping • 6.2 Factoring Trinomials of the Form ax2 + bx + c • 6.3 Factor Trinomials of the Form ax2 + bx + c,a ≠1

  2. Section 6.1 Objectives 1 Find the Greatest Common Factor of Two or More Expressions 2 Factor Out the Greatest Common Factor in Polynomials 3 Factor Polynomials by Grouping

  3. factors Factors 4· 9 = 36 3(x + 2) = 3x + 6 (2x – 7)(3x + 5) = 6x2 – 11x – 35 The expressions on the left side are called factors of the expression on the right side. To factor a polynomial means to write the polynomial as a product of two or more polynomials. The greatest common factor (GCF) of a list of algebraic expressions is the largest expression that divides evenly into all the expressions.

  4. The common factors are 2 and 2. Greatest Common Factor How to Find the Greatest Common Factor of a List of Numbers Step 1: Write each number as a product of prime factors. Step 2: Determine the common prime factors. Step 3: Find the product of the common factors found in Step 2. This number is the GCF. Example: Find the GCF of 16 and 20. 16 = 2 · 2 · 2 · 2 20 = 2 · 2 · 5 The GCF is 2 · 2 = 4.

  5. The common factors are 3 and 5. Greatest Common Factor Example: Find the GCF of 60, 75, and 135. 60 = 2 · 2 · 3 · 5 75 = 3 · 5 · 5 135 = 3 · 3 · 3 · 5 The GCF is 3 · 5 = 15.

  6. The GCF as a Binomial Example: Find the GCF of 6(x – y) and 15(x – y)3. 6(x – y) = 2 · 3 · (x – y) 15(x – y)3 = 3 · 5 · (x – y) · (x – y) · (x – y) The GCF is 3 · (x – y) = 3(x – y).

  7. Finding the GCF Steps to Find the Greatest Common Factor Step 1: Find the GCF of the coefficients of each variable factor. Step 2: For each variable factor common to all terms, determine the smallest exponent that the variable factor is raised to. Step 3: Find the product of the common factors found in Steps 1 and 2. This expression is the GCF.

  8. Factoring Polynomials Steps to Factor a Polynomial Using the GCF Step 1: Identify the GCF of the terms that make up the polynomial. Step 2: Rewrite each term as the product of the GCF and the remaining factor. Step 3: Use the Distributive Property “in reverse” to factor out the GCF. Step 4: Check using the Distributive Property.

  9. Factoring Polynomials Example: Factor the trinomial 36a6 + 45a4 – 18a2 by factoring out the GCF. Step 1: Find the GCF. GCF = 9a2 Step 2: Rewrite each term as the product of the GCF and the remaining term. 36a6 + 45a4 – 18a2 = 9a2· 4a4 + 9a2 · 5a2 –9a2 · 2 Step 3: Factor out the GCF. 36a6 + 45a4 – 18a2 = 9a2(4a4 + 5a2 – 2) Step 4: Check. 9a2(4a4 + 5a2 – 2) =36a6 + 45a4 – 18a2 

  10. Factoring Out a Negative Number Example: Factor – 3x6 + 9x4 – 18x by factoring out the GCF: Step 1: Find the GCF. GCF = – 3x Step 2: Rewrite each term as the product of the GCF and the remaining term. – 3x6 + 9x4 – 18x = – 3x· x5 + (– 3x)(– 3x3) + (– 3x) · 6 Step 3: Factor out the GCF. – 3x6 + 9x4 – 18x = – 3x(x5 – 3x3 + 6) Step 4: Check. – 3x(x5 – 3x3 + 6) =– 3x6 + 9x4 – 18x 

  11. GCF = 3x + y Factoring Out a Binomial Example: Factor out the greatest common binomial factor: 6(3x + y) – z(3x + y) 6(3x + y) – z(3x + y) = (3x + y)(6 – z) Check: (3x + y)(6 – z) = 6(3x + y) – z(3x + y) 

  12. Factoring by Grouping Steps to Factor a Polynomial by Grouping Step 1: Group the terms with common factors. Step 2: In each grouping, factor out the greatest common factor. Step 3: If the remaining factor in each grouping is the same, factor it out. Step 4: Check your work by finding the product of the factors.

  13. x is the common factor. 3 is the common factor. Factor out the x from the first two terms. Factor out the 3 from the last two terms. These two factors need to be the same. Factoring by Grouping Example: Factor by grouping: x2 + 7x + 3x + 21 x2 + 7x + 3x + 21 x2 + 7x + 3x + 21 = x(x + 7) + 3(x + 7) = (x + 3)(x + 7) Factor out the common factor x + 7. (x + 3)(x + 7) = x2 + 7x + 3x + 21 Check: 

  14. Factoring by Grouping Example: Factor by grouping: xy – 4x – 3y + 12 Step 1: Group the terms with the common factors. (xy – 4x) + (– 3y + 12) Step 2: Factor out the common factor in each group. xy – 4x – 3y + 12 = x(y – 4) + (– 3)(y – 4) Step 3: Factor out the common factor that remains. = (x – 3)(y – 4) xy – 4x – 3y + 12 = (x – 3)(y – 4) Step 4: Check. (x – 3)(y – 4) = xy – 4x – 3y + 12 

  15. Chapter 6 Factoring Polynomials Section 2 Factoring Trinomials of the Formax2 + bx + c

  16. Section 6.2 Objectives 1 Factor Trinomials of the Form ax2 + bx + c 2 Factor Out the GCF, then Factor ax2 + bx + c

  17. leading coefficient Quadratic Trinomials A quadratic trinomial is a polynomial of the form ax2 + bx + c, a 0 where a represents the coefficient of the squared (second degree) term, b represents the coefficient of the linear (first degree) term and c represents the constant. 3x2 + 4x + 7 8a2 – 24a – 10 6c2 + c – 25 When the trinomial is written in standard form (or descending order of degree), the coefficient of the squared term is called the leading coefficient.

  18. Trinomials of the Form x2 + bx + c Factoring a Trinomial of the Form x2 + bx + c Step 1: Find the pair of integers whose product is c and whose sum is b. That is, determine m and n such that mn = c and m + n = b. Step 2: Write x2+ bx + c = (x + m)(x + n). Step 3: Check your work by multiplying the binomials. The coefficient of x is the sum of the two numbers. x2 – 11x + 24 = (x – 3)(x – 8) The last term is the product of the two numbers.

  19. Trinomials of the Form x2 + bx + c Example: Factor: x2 + 8x + 15 x2 + 8x + 15 = (x + ?)(x + ?) 3 5 ___ × ___ = 15 Find two numbers that we can multiply together to get 15 and add together to get 8. 3 5 ___ + ___ = 8 x2 + 8x + 15 = (x + 3)(x + 5) Check:  (x + 3)(x + 5) = x2 + 8x + 15

  20. Trinomials of the Form x2 + bx + c, c < 0 Example: Factor: x2 + x – 42 x2 + x – 42 = (x + ?)(x– ?) One factor will be positive and one will be negative. ( 6) 7 ____ × ____ =  42 Find two numbers that we can multiply together to get – 42 and add together to get 1. ( 6) 7 ____ + ____ = 1 x2 + x – 42 = (x + 7)(x6) Check:  (x + 7)(x – 6 ) = x2 + x – 42

  21. Trinomials of the Form x2 + bx + c The following table summarizes the four forms for factoring a quadratic trinomial in the form x2 + bx + c.

  22. Neither of these sums produce the correct middle term. Prime Polynomials A polynomial that cannot be written as the product of two other polynomials (other than 1 or – 1) is said to be a prime polynomial. Example: Factor: 5x2 x 2 The Factors of 2 are: The Factors of 5 are: 1 and 5 1 and 2 Possible Factors: Middle Term: (x – 1)(5x + 2) – 3x – 9x (x – 2)(5x + 1) The polynomial 5x2 x 2 is prime.

  23. Trinomials with a Common Factor Example: Factor: 2x2 – 32x + 96 2x2 – 32x + 96 = 2(x2 – 16x + 48) The common factor of 2 can be factored out. x2 – 16x + 48 = (x–?)(x– ?) ( 4) ( 12) ____ × ____ = 48 Find two numbers that we can multiply together to get 48 and add together to get – 16. ( 4) ( 12) ____ + ____ = – 16 2x2 – 32x + 96 = 2(x– 12)(x4) Check: 2(x – 12)(x 4) = 2(x2 – 16x + 48)  = 2x2 – 32x + 96

  24. Negative Leading Coefficients Example: Factor: – x2 – 12x – 36 It is easier to factor out a GCF of – 1. – x2 – 12x – 36 = – 1(x2 + 12x + 36) – 1(x2 + 12x + 36) = – 1(x + ?)(x + ?) 6 6 ___ × ___ = 36 Find two numbers that we can multiply together to get 36 and add together to get 12. 6 6 ___ + ___ = 12 – 1(x2 + 12x + 36) = – 1(x + 6)(x + 6) = –(x + 6)2 Check: –(x + 6)2 =– 1(x + 6)(x + 6)  =– x2 – 12x – 36

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

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

  27. ( __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

  28. 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: 

  29. 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: 

  30. 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: 

  31. 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.

  32. 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 

  33. 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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