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Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.2 – Slide 1 Chapter 6 Factoring and Applications Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.2 – Slide 2 6.2 Factoring Trinomials Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.2 – Slide 3 6.2 Factoring Trinomials Objectives 1. 2. Factors trinomials with a coefficient of 1 for the squared term. Factor trinomials after factoring out the greatest common factor. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.2 – Slide 4 6.2 Factoring Trinomials Factoring Trinomials with a Coefficient of 1 for Squared Term Example 1 Factor x2 + 8x + 12. Look for two integers whose product is 12 and whose sum is 8. Only positive integers are needed since all signs in x2 + 8x + 12 are positive. From the list, 6 and 2 are the required Factors Sums of integers. Thus, of 12 Factors x2 + 8x + 12 = (x + 6)(x + 2) 12, 1 13 Check: 4, 3 7 2 + 2x + 6x + 12 (x + 6)(x + 2) = x 6, 2 8 = x2 + 8x + 12 Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.2 – Slide 5 6.2 Factoring Trinomials Factoring Trinomials with a Coefficient of 1 for Squared Term Note In Example 1, the answer (x + 2)(x + 6) also could have been written (x + 6)(x + 2). Because of the commutative property of multiplication, the order of the factors does not matter. Always check by multiplying. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.2 – Slide 6 6.2 Factoring Trinomials Factoring Trinomials with a Coefficient of 1 for Squared Term Example 2 Factor x2 – 9x + 18. Find two integers whose product is 18 and whose sum is – 9. Since the numbers we are looking for have a positive product and a negative sum, we consider only pairs of negative integers. Factors Sums of of 18 Factors –18, –1 –19 –9, –2 –6, –3 –11 –9 The required integers are –6 and –3, so x2 – 9x + 18 = (x – 6)(x – 3) Check: (x – 6)(x – 3) = x2 – 3x – 6x + 18 = x2 – 9x + 18 Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.2 – Slide 7 6.2 Factoring Trinomials Factoring Trinomials with a Coefficient of 1 for Squared Term Example 3 Factor x2 – 3x – 10. This time, we need to consider positive and negative integers whose product is –10 and whose sum is –3. Factors Sums of of –10 Factors 10, –1 9 –10, 1 –9 5, –2 3 –5, 2 –3 Here –5 and 2 are the required integers. x2 – 3x – 10 = (x – 5)(x + 2) Check: (x – 5)(x + 2) = x2 + 2x – 5x – 10 Copyright © 2010 Pearson Education, Inc. All rights reserved. = x2 – 3x – 10 6.2 – Slide 8 6.2 Factoring Trinomials Factoring Trinomials with a Coefficient of 1 for Squared Term Example 4 Factor x2 + 3x + 9. Both factors must be positive to give a positive product and a positive sum. We need to consider only positive integers. Factors Sums of of 9 Factors 9, 1 11 3, 3 6 None of the pairs of integers has a sum of 3. Therefore, the trinomial cannot be factored using only integers. This is a prime polynomial. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.2 – Slide 9 6.2 Factoring Trinomials Factoring Trinomials with a Coefficient of 1 for Squared Term Factoring x2 + bx + c Find two integers whose product is c and whose sum is b. 1. Both integers must be positive if b and c are positive. 2. Both integers must be negative if c is positive and b is negative. 3. One integer must be positive and one must be negative if c is negative. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.2 – Slide 10 6.2 Factoring Trinomials Factoring Trinomials with a Coefficient of 1 for Squared Term Example 5 Factor x2 + 6xy – 7y2. Though this trinomial has two variables, the process for factoring it still works the same. Here 7y and –y will work. So, Factors Sums of of –7y2 Factors –7y, 1y –6y 7y, –1y 6y x2 + 6xy – 7y2 = (x + 7y)(x – y) Check: (x + 7y)(x – y) = x2 – xy + 7xy – 7y2 = x2 + 6xy – 7y2 Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.2 – Slide 11 6.2 Factoring Trinomials Factoring Trinomials after Factoring out the GCF Example 6 Factor 3a7 – 30a6 – 33a5. First, factor out the greatest common factor, 3a5. Then, factor the trinomial as usual. 3a7 – 30a6 – 33a5 = 3a5(a2 – 10a – 11) = 3a5(a – 11)(a + 1) Check: 3a5(a – 11)(a + 1) = 3a5(a2 – 10a – 11) = 3a7 – 30a6 – 33a5 Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.2 – Slide 12