# x 4-1 = (x²)² - (1)² = (x² + 1) (x² - 1) = (x² + 1) (x - 1) (x + 1)

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1 Factoring Polynomials EXAMPLES STEP 1 : Greatest Common Factor GCF Factor out the greatest common factor. 6x³ + 12x²y = 6x² (x + 2y) 5x - 5 = 5 (x - 1) 7x² + 2y² = 1 (7x² + 2y²) 2x (x - 3) - (x - 3) = (x - 3) (2x - 1) STEP 2 : Count Terms Two Terms 1) Difference of two squares. (Binomials) 4-9x² = (2)² - (3x)² a² - b² = (a + b) (a - b) =(2 + 3x) (2-3x) or (a - b) (a + b) x 4-1 = (x²)² - (1)² = (x² + 1) (x² - 1) = (x² + 1) (x - 1) (x + 1) 2) Difference of two cubes. 8x³ - 27 = (2x)³ - (3)³ a³ - b³ = (a - b) (a² + ab + b²) = (2x - 3) ( (2x)² + (2x)(3) + (3)² ) = (2x - 3) (4x² + 6x + 9) y³ - 1 = (y)³ - (1)³ = (y - 1) (y² + y + 1) 3) Sum of two cubes 8x³ + 27 = (2x)³ + (3)³ a³ + b³ = (a + b) (a² - ab + b²) = (2x + 3) ( (2x)² - (2x)(3) + (3)² ) = (2x + 3) ( 4x² - 6x + 9) y³ + 1 = (y)³ + (1)³ = (y + 1) (y² - y + 1) Three Terms 1) Leading coefficient is 1. (Trinomials) x² - 5x - 24 x² + bx + c Find two integers whose product = -24 and Find two integers whose product = c whose sum = -5. and whose sum = b -8 3 = -24 and = -5 so, x² - 5x - 24 = (x - 8) (x + 3) x² + x - 12 Find two integers whose product = -12 and whose sum = 1. 4 (-3) = -12 and 4 + (-3) = 1, so, x² + x - 12 = (x + 4) (x - 3). x² - 5x + 6 Find two integers whose product = 6 and whose sum = -5. (-3) (-2) = 6 and (-3) + (-2) = -5, so, x² - 5x + 6 = (x - 3) (x - 2). Reprint with permission only: Chandler - Gilbert Community College Learning Center 1

2 2) Leading coefficient is -1. -x² + 5x + 24 = -(x² - 5x - 24) -x² + bx + c factor out -1 first. = -(x - 8) (x + 3) -1 (x² - bx - c) then factor (x² - bx - c) using step 1 above. -x² + 5x - 6 = -(x² - 5x + 6) = -(x - 3) (x - 2) 3) Leading coefficient is not 1 or x² + 3x - 10 ax² + bx + c, a 1, -1 Zeros (by calculator) are 2/3 and -5/6, so i) you may use your calculator to check this polynomial is factorable. The factors if this is factorable. Find the zeros. are (3x - 2) and (6x + 5). So, If the zeros are integers or rational 18x² + 3x - 10 = (3x - 2) (6x + 5). numbers, then the given trinomial is factorable. Otherwise, the trinomial is prime. On the other hand, 18x² + 3x + 10 has imaginary zeros and, 5x² - 10x + 3 has irrational zeros so both are prime. xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx ii) ac - method: ax² + bx + c 18x² + 3x - 10 Find the product of the coefficients a and c. Then find two integers whose product ac = (18) (-10) = -180 = ac and whose sum = b. Find two integers whose product = -180 and Rewrite the middle term replacing b by whose sum = 3. the sum of the two integers. Factor the (15) (-12) = -180 resulting four-term polynomial by grouping (-12) = 3 so, 18x² + 3x - 10 (See section below = 18x² + 15x - 12x - 10 on factoring four-term polynomials.) = 3x (6x + 5) - 2 (6x + 5) by grouping = (6x + 5) (3x - 2) You may organize the terms in a box and find the greatest common factor of each row and each column, factoring out leading minus signs. l l l 18x² l +15x l -12x l -10 l 6x l +5 3x l 18x² l +15x -2 l -12x l -10 so, 18x² + 3x - 10 = (3x - 2) (6x + 5) Reprint with permission only: Chandler - Gilbert Community College Learning Center 2

3 iii) Trial and error method ax² + bx + c 12x² + 43x + 35 Write possible factorizations of a and of c. 12 = = 35 1 = 6 2 = 5 7 Find the sum of the outer and inner = 4 3 products. We need this sum to equal bx. (12x + 35) (x + 1) (12x + 1) (x + 35) If b and c are both positive, the factors 35x x have only positive coefficients. 12x 420x 35x + 12x = 47x 43x x + 420x = 421x 43x (12x + 5) (x + 7) (12x + 7) (x + 5) 5x 7x 84x 60x 5x + 84x = 89x 43x 7x + 60x = 67x 43x (6x + 5) (2x + 7) (6x + 7) (2x + 5) 10x 14x 42x 30x 10x + 42x = 52x 43x 14x + 30x = 44x 43x (4x + 7) (3x + 5) (4x + 5) (3x + 7) 21x 15x 20x 28x 21x + 20x = 41x 43x 15x + 28x = 43x so, 12x² + 43x + 35 = (4x + 5) (3x + 7) xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx If b is positive and c is negative, one of the 18x² + 3x - 10 factors of c must be negative and the sum of the outside and inside products must be 18 = = = 10 (-1) positive. = 9 2 = -5 2 = 5 (-2) = and or 10 (-1) most likely will not work because 3 will not be a sum. (9x - 5) (2x + 2) (9x - 2) (2x + 5) -10x -4x 18x 45x -10x + 18x = 8x 3x -4x + 45x = 41x 3x (6x + 5) (3x - 2) 15x -12x 15x + -12x = 3x so, 18x² + 3x - 10 = (6x + 5) (3x - 2) Reprint with permission only: Chandler - Gilbert Community College Learning Center 3

4 If b and c are both negative, one of the 12x² - 13x - 35 factors of c must be negative and the sum of the outside and inside products must 12 = = = 35 (-1) be negative. = 6 2 = -5 7 = 5 (-7) = 4 3 (6x - 5) (2x + 7) -10x 42x -10x + 42x = 32x -13x (4x - 5) (3x + 7) (4x + 5) (3x - 7) -15x 15x 28x -28x -15x + 28x = 13x -13x 15x - 28x = -13x so, 12x² - 13x - 35 = (4x + 5) (3x - 7) Four Terms If there are four terms, consider factoring by 18x² + 3x - 10 grouping the terms into two groups of two terms each. Factor the GCF from each 18x² + 15x - 12x - 10 group. If there is then a common binomial factor, factor it out. = 3x (6x + 5) - 2(6x + 5) = (6x + 5) (3x - 2) You may organize the terms in a box and find the greatest common factor of each row and each column, factoring out leading minus signs. l l l 18x² l +15x l -12x l -10 l 6x l +5 3x l 18x² l +15x -2 l -12x l -10 so, 18x² + 3x - 10 = (3x - 2) (6x + 5) 12x² - 28x + 15x - 35 = 4x(3x - 7) + 5(3x - 7) = (3x - 7) (4x + 5) OR l 3x l -7 4x l 12x² l -28x +5 l +15x l -35 Reprint with permission only: Chandler - Gilbert Community College Learning Center 4

5 3x³ + 7x² - 15x - 35 = x² (3x + 7) - 5(3x + 7) = (3x + 7) (x² - 5) OR l 3x l -7 x² l 3x³ l +7x² -5 l -15xl -35 STEP 3 : Factor Completely Examine each non-monomial factor x 4-81 (Difference of 2 squares) resulting from the above steps and check if = (x² + 9) (x² - 9) (x² - 9 is also Difference of 2 squares) it is further factorable = (x² + 9) (x + 3) (x - 3) Repeat this step until each 5x non-monomial factor is prime. = 5 (x 4-81) = 5 (x² + 9) (x² - 9) = 5 (x² + 9) (x - 3) (x + 3) x 12 - y 6 = (x 6 )² - (y³)² = (x 6 - y 3 ) (x 6 + y 3 ) = ((x²)³ - y³) ((x²)³ + y³) = (x² - y) (x 4 + x²y + y²) (x² + y) (x 4 - x²y + y²) 48x x³ - 30x² = 6x² (8x² + 18x - 5) = 6x² (8x² + 20x - 2x - 5) = 6x² (2x + 5) (4x - 1) where l 2x l +5 4x l 8x² l +20x -1 l -2x l -5 Note: Calculators with symbolic manipulation features can factor polynomials directly. Regardless of the method used to factor polynomials, the importance of factors is their use in solving problems. Reprint with permission only: Chandler - Gilbert Community College Learning Center 5

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### CONVERT QUADRATIC FUNCTIONS FROM ONE FORM TO ANOTHER (Standard Form <==> Intercept Form <==> Vertex Form) (By Nghi H Nguyen Dec 08, 2014) CONVERT QUADRATIC FUNCTIONS FROM ONE FORM TO ANOTHER (Standard Form Intercept Form Vertex Form) (By Nghi H Nguyen Dec 08, 2014) 1. THE QUADRATIC FUNCTION IN INTERCEPT FORM The graph of the quadratic Alum Rock Elementary Union School District Algebra I Study Guide for Benchmark III Name Date Adding and Subtracting Polynomials Algebra Standard 10.0 A polynomial is a sum of one ore more monomials. Polynomial CHAPTER 6 Factoring Polynomials 6.1 The Greatest Common Factor and Factoring by Grouping 6. Factoring Trinomials of the Form b c 6.3 Factoring Trinomials of the Form a b c and Perfect Square Trinomials