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M098 Carson Elementary and Intermediate Algebra 3e Section 6.3 Objectives 1. 2. 2 Factor trinomials of the form ax + bx + c, where, by trial. 2 Factor trinomials of the form ax + bx + c, where a ≠ 1, by grouping. Vocabulary Prior Knowledge Factoring Sign Clues Factoring by Grouping New Concepts 2 1. Factoring trinomials of the ax + bx + c where a ≠ 1. 2 The general form for a trinomial is ax + bx + c. The coefficient of the squared term is referred to as a, the coefficient of the middle term is b and the constant is c. Some of the clues that were presented in the last section work only when a is 1. The problems become more difficult when a is not 1. The book shows two methods for factoring these types of problems. Objective 1 shows the “trial and error” or “guess and check” method. Trial and error is preferred by many students and instructors because it is more of a mental process and is faster. Some people are gifted at seeing these combinations of numbers and others struggle with them. The more practice you have, the easier they become. Objective 2 shows the factoring by grouping method. This method rewrites each trinomial as a 4-term polynomial and then uses factoring by grouping. There is a third method that was shared on the Internet a few years ago that also works well. It is related to factoring by grouping. We call this Factoring the Illegal Way. Watch the animation to see how the Illegal Factoring Method works. The Illegal Method: Factor: 6x 2 7x 3 This is difficult to factor because the coefficient of the squared term (a) is not 1. To remove the 6, multiply it with the c term (-3). 6x 2 7x 3 The new trinomial is x 2 7x 18 This trinomial is easily factored as x 9x 2 . An “illegal move” was committed so now it must be undone. Since 6 was multiplied in the first step, to “undo” the illegal move, divide each constant in the binomials by 6. 9 2 x x 6 6 This is factored form but integers are preferred in the binomials rather than fractions. To remove the fractions, first reduce the fractions to lowest terms. 3 1 x x 2 3 V. Zabrocki 2011 page 1 M098 Carson Elementary and Intermediate Algebra 3e Section 6.3 Now take the denominator of each fraction and slide it in front of the x in the binomial, making the denominator the coefficient of x. 3 1 x x 2 3 The Illegal Move does not work if a is a negative number. Factor out -1 and then proceed. 2x 33x 1 Always foil to check your factors. 2x 33x 1 6x 2 7x 3 Do not forget to undo the illegal move. You must do both steps!!! Example 1: 2 Factor 2x + 7x + 3 Factoring by Grouping Multiply a · c and look for factors of that number that add (because of the +3) to equal 7. Illegal Method Multiply a · c and rewrite the trinomial. 2 Guess & Check Since 2 and 3 are both prime numbers, there is only one way to factor each of them. x + 7x + 6 Illegal Move - ·2 2·1=2 3·1=3 2·3=6 Factor using the sign clues: 6=1·6 6 = 2· 3 (x + 1)(x + 6) Rewrite as a 4-term polynomial replacing the middle term (7x) with (1x + 6x). 2 2x +1x + 6x + 3 Group the first and last terms. 2 Notice when this is foiled it does NOT equal the original trinomial. That’s because we need to undo the Illegal Move. The Illegal Move was to multiply by 2 so to undo the Illegal Move, divide the constants by 2. (2x + 1x) + (6x + 3) Pull out the GCF from each binomial. x(2x + 1) + 3(2x + 1) Factor out the binomial GCF. (2x + 1)(x + 3) Always foil to be sure the factors are correct. 2 1 6 x x 2 2 Reduce the fractions if you can and slide the remaining denominator in front of the variable. 1 x x 3 2 (2x + 1)(x + 3) Start with the factors of 2 and place them in the binomials. (2x )( x ) The sign clues tell us the signs are the same and both positive. (2x + )( x + ) Now for the guess and check. The last terms must be either 3 and 1 or 1 and 3. Try one combination and foil to see if it works. (2x + 3)( x + 1) 2 2 2x + 2x + 3x + 3 = 2x + 5x + 3 The check didn’t work so switch the 3 and 1 and check again by foiling. (2x + 1)( x + 3) 2 2 2x + 6x + 1x + 3 = 2x + 7x + 3 (2x + 1)(x + 3) = 2x + 7x + 3 Always foil to be sure the factors are correct. It works. 2 (2x + 1)(x + 3) = 2x + 7x + 3 V. Zabrocki 2011 page 2 M098 Carson Elementary and Intermediate Algebra 3e Section 6.3 Example 2: Do each of the following problems using all three methods until you decide which one works the best for you. 2 3m – 10m + 3 (3m – 1)(m – 3) 2 (2y + 3)(2y + 1) 2 (4u + 3)(u – 5) 2 (3z + 5)(z – 6) 4y + 8y + 3 4u – 17u – 15 3z – 13z – 30 2 2 9x + 9xy – 10y 2 6u + 3u – 7 (3x – 2y)(3x + 5y) prime 2. The first step in factoring should always be to remove the GCF. Example 3: Factor. Remove the GCF first. 2 6a – 20a + 16 2 20xy – 50xy + 30x V. Zabrocki 2011 2(3a – 4)(a – 2) 10x(y – 1)(2y – 3) page 3