Section 6.5 A General Factoring Strategy

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1 Difference of Two Squares: a 2 b 2 = (a + b)(a b) NOTE: Sum of Two Squares, a 2 b 2, is not factorable Sum and Differences of Two Cubes: a 3 + b 3 = (a + b)(a 2 ab + b 2 ) a 3 b 3 = (a b)(a 2 + ab + b 2 ) Perfect Square Trinomials: a 2 + 2ab + b 2 = (a + b) 2 a 2 2ab + b 2 = (a b) 2 Strategy for Factoring Polynomials Step 1: Factor out GCF, if possible Step 2: Determine the number of terms in the polynomial as follows: If the given polynomial is a Binomial, factoring by one of the following 1. a 2 b 2 = (a + b)(a b) 2. a 3 + b 3 = (a + b)(a 2 ab + b 2 ) 3. a 3 b 3 = (a b)(a 2 + ab + b 2 ) If the given polynomial is a Trinomial, factoring by ac-method If the given polynomial has 4 or more terms, factoring by Grouping Exercises (Solution 1) Step 1: Factor out GCF. GCF = 2x 2x 2 + 8x = 2x( x + 4) Since 2x( x + 4) is the combination of two linear, the polynomial is factored completely. Factoring is done! The answer is 2x( x + 4) or 2x(x 4) Cheon-Sig Lee Page 1

2 (Solution 2) Since two terms, power 2, and difference, we are using a 2 b 2 for factoring a 2 b 2 = (a + b)(a b) Factoring by a 2 b 2 4x 2 25 = 2 2 x = (2x) Step 2: Find a and b, then factor out with a 2 b 2 (2x) a = 2x, b = 5 = (2x + 5)(2x 5) Two linear combination, so factoring is done (Solution 3) Since two terms, power 3, and difference, we are using a 3 b 3 for factoring a 3 b 3 = (a b)(a 2 + ab + b 2 ) Factoring by a 3 b 3 216y 3 1 = 6 3 y = (6y) Step 2: Find a and b, then factor out with a 3 b 3 (6y) a = 6y, b = 1 = (6y 1)[(6y) 2 + (6y)(1) ] = (6y 1)(36y 2 + 6y + 1) (Solution 4) 2x + 2y + x 2 + xy is four terms, so By Grouping Factoring by Grouping Step 1: Group first two terms and last two terms (2x + 2y) + (x 2 + xy) Step 2: Factor out GCF from each group GCF for the first group is 2 GCF for the second group is x (2x + 2y) + (x 2 + xy) = 2(x + y) + x(x + y) = (x + y)(2 + x) Cheon-Sig Lee Page 2

3 (Solution 5) Ignore FOIL multiplication Step 1: Factor out GCF. GCF is 1, so DONE! Step 2: Find a, b, c a = 8, b = 22, c = 15 Step 3: Find ac ac = 8 15 = 120 Step 4: Find positive factors of ac whose products is ac and whose sum is b Positive factors of 40: (1 120), (2 60), (3 40), (4 30) (5 24), (6 20), (8 15), (10 12) Since ac is positive, there are two case; Case 1: Both factors are positive Case 2: Both factors are negative Since b is negative, both factors are negative Thus, possible factors are ( 1) ( 120), ( 2) ( 60), ( 3) ( 40) ( 4) ( 30), ( 5) ( 24), ( 6) ( 20) ( 8) ( 15), ( 10) ( 12) Step 5: Rewrite the middle term, bx, using the two numbers found in step 4 as coefficients. Then factoring by grouping. 8x 2 22x + 15 = 8x 2 10x 12x + 15 = (8x 2 10x) (12x 15) = 2x(4x 5) 3(4x 5) = (4x 5)(2x 3) Ignore FOIL multiplication (Solution 6) Step 1: Factor out GCF. GCF = 3x 3x x = 3x(x 2 + 4) (x 2 + 4) = x Sum of Two Squares The sum of two squares is not factorable So, factoring is done The answer is 3x(x 2 + 4) Cheon-Sig Lee Page 3

4 (Solution 7) Step 1: Factor out GCF. GCF = 5x 10x 2 + 5x = 5x(2x + 1) Since 5x(2x + 1) is the combination of two linear, the polynomial is factored completely. Factoring is done! The answer is 5x(2x + 1) (Solution 8) Step 1: Factor out GCF. GCF is 1, so DONE! Step 2: Find a, b, c a = 1, b = 5, c = 25 Step 3: Find ac ac = 1 25 = 25 Step 4: Find positive factors of ac whose products is ac and whose sum is b Positive factors of 25: (1 25), (5 5) Since ac is positive, there are two case; Case 1: Both factors are positive Case 2: Both factors are negative Since b is negative, both factors are negative Thus, possible factors are ( 1) ( 25), ( 5) ( 5) ( 1) + ( 25) = 26 No combination gives the ( 5) + ( 5) = 10 middle term 5. Thus, the polynomial is prime (Solution 9) Step 1: Factor out GCF. GCF = 3x 3x 3 27x = 3x(x 2 9) Step 2: Determine the number of terms: x 2 9 has two terms Since two terms, power 2, and difference, we are using a 2 b 2 for factoring a 2 b 2 = (a + b)(a b) Factoring by a 2 b 2 3x(x 2 9) = 3x(x ) Step 2: Find a and b, then factor out with a 2 b 2 3x(x ) a = x, b = 3 = 3x(x + 3)(x 3) Three linear combination, so factoring is done Cheon-Sig Lee Page 4

5 (Solution 9) Since two terms, power 2, and sum, it is prime x = x sum of two squares Remember! Sum of Two Squares is not factorable Cheon-Sig Lee Page 5