Copyright © 2012 Pearson Education, Inc. 6.3 Trinomials of the Type ax2 + bx + c ■ Factoring Trinomials of the Type ax2 + bx + c ■ Equations and Functions Copyright © 2012 Pearson Education, Inc.

Copyright © 2012 Pearson Education, Inc. METHOD 2: The ac-Method Copyright © 2012 Pearson Education, Inc.

Copyright © 2012 Pearson Education, Inc. To Factor ax2 + bx + c, Using the ac-Method 1. Factor out the largest common factor, if one exists. 2. Multiply the leading coefficient a and the constant c. 3. Find a pair of factors ac whose sum is b. 4. Rewrite the middle term as a sum or a difference using the factors found in step (3). 5. Factor by grouping. 6. Include any common factor from step (1) and check by multiplying. Copyright © 2012 Pearson Education, Inc.

Copyright © 2012 Pearson Education, Inc. Example Factor 4x2  5x  6 Solution 1. First, we note that there is no common factor (other than 1 or 1). 2. Multiply the leading coefficient, 4 and the constant, 6: (4)(6) = 24. 3. Try to factor 24 so that the sum of the factors is 5: Copyright © 2012 Pearson Education, Inc.

Copyright © 2012 Pearson Education, Inc. continued Factor 4x2  5x  6 3. Pairs of Factors of 24 Sums of Factors 1, 24 23 1, 24 23 2, 12 10 2, 12 10 3, 8 5 3, 8 5 4, 6 2 4, 6 2 We would normally stop listing pairs of factors once we have found the one we are after. Copyright © 2012 Pearson Education, Inc.

Copyright © 2012 Pearson Education, Inc. continued Factor 4x2  5x  6 4. Split 5x using the results of step (3): 5x = 8x + 3x. 5. Factor by grouping: 4x2  5x  6 = 4x2  8x + 3x  6 = 4x(x  2) + 3(x  2) = (x  2)(4x + 3) We check the solution by multiplying or using a table. Check: (x  2)(4x + 3) = 4x2 + 3x  8x  6 = 4x2  5x  6 The factorization of 4x2  5x  6 is (x  2)(4x + 3). Copyright © 2012 Pearson Education, Inc.

Copyright © 2012 Pearson Education, Inc. Example Factor 8x3 + 10x2  12x Solution 1. We factor out the greatest common factor, 2x: 8x3 + 10x2  12x = 2x(4x2 + 5x  6) 2. To factor 4x2 + 5x  6 by grouping, we multiply the leading coefficient, 4 and the constant term (6): 4(6) = 24. 3. We next look for pairs of factors of 24 whose sum is 5. Pairs of Factors of 24 Sums of Factors 3, 8 5 3, 8 5 Copyright © 2012 Pearson Education, Inc.

Copyright © 2012 Pearson Education, Inc. continued Factor 8x3 + 10x2  12x 4. We then rewrite the middle term: 4x2 + 5x  6 using 5x = 3x + 8x 5. Factor by grouping: 4x2 + 5x  6 = 4x2  3x + 8x  6 = x(4x  3) + 2(4x  3) = (x + 2)(4x  3) 6. The factorization of the original trinomial 8x3 + 10x2  12x = 2x(x + 2)(4x  3). Copyright © 2012 Pearson Education, Inc.

Equations and Functions We now use our new factoring skills to solve a polynomial equation. We factor a polynomial expression and use the principle of zero products to solve the equation. Copyright © 2012 Pearson Education, Inc.

Copyright © 2012 Pearson Education, Inc. Example Solve: 2x5 – 7x4 + 3x3 = 0. Solution--Algebraic We note that the polynomial has degree 5, so there will be at most 5 solutions of the equation. 2x5 – 7x4 + 3x3 = 0 x3(2x – 1)(x – 3) = 0 Factoring x3 = 0 or 2x – 1 = 0 or x – 3 = 0 x = 0 or x = ½ or x = 3 The solutions are 0, ½ and 3. Copyright © 2012 Pearson Education, Inc.

continued--Graphical We find the x-intercepts of the function 2x5 – 7x4 + 3x3 using the ZERO option of the CALC menu. (0, 0) (0.5, 0) (3, 0) Copyright © 2012 Pearson Education, Inc.

Copyright © 2012 Pearson Education, Inc. Example Find the domain of F if F(x) = Solution The domain of F is the set of all values for which the function is a real number. Since division by 0 is undefined, we exclude any x-value for which the denominator is 0. x2 + 2x – 8 = 0 (x – 2)(x + 4) = 0 x – 2 = 0 or x + 4 = 0 x = 2 or x = –4 These are the values to exclude. The domain of F is {x|x is a real number and x 2 and x  4}. Copyright © 2012 Pearson Education, Inc.