1. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §5.6 Factoring Strategies

2. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §5.5 → Factoring: TriNomials, Special Forms  Any QUESTIONS About HomeWork §5.5 → HW-19 5.5 MTH 55

3. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 3 Bruce Mayer, PE Chabot College Mathematics To Factor a Polynomial A.Always look for a common factor first. If there is one, factor out the Greatest Common Factor (GCF). Be sure to include it in your final answer. B.Then look at the number of terms TWO Terms: If you have a Difference of SQUARES, factor accordingly: A 2 − B 2 = (A − B)(A + B)

4. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 4 Bruce Mayer, PE Chabot College Mathematics To Factor a Polynomial CUBESTWO Terms: If you have a SUM of CUBES, factor accordingly: A 3 + B 3 = (A + B)(A 2 − AB + B 2 ) CUBESTWO Terms: If you have a DIFFERENCE of CUBES, factor accordingly: A 3 − B 3 = (A − B)(A 2 + AB + B 2 ) THREE Terms: If the trinomial is a perfect-square trinomial, factor accordingly: A 2 + 2AB + B 2 = (A + B) 2 or A 2 – 2AB + B 2 = (A – B) 2.

5. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 5 Bruce Mayer, PE Chabot College Mathematics To Factor a Polynomial THREE Terms: If it is not a perfect- square trinomial, try using FOIL Guessing FOUR Terms: Try factoring by grouping C.Always factor completely: When a factor can itself be factored, be sure to factor it. Remember that some polynomials, like A 2 + B 2, are PRIME D.CHECK by Multiplying the Factors

6. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 6 Bruce Mayer, PE Chabot College Mathematics Choosing the Right Method  Example: Factor 5t 4 − 3125  SOLUTION A.Look for a common factor: 5t 4 − 3125 = 5(t 4 − 625). B.The factor t 4 − 625 is a diff of squares: (t 2 ) 2 − 25 2. We factor it, being careful to rewrite the 5 from step (A): 5(t 4 − 625) = 5(t 2 − 25)(t 2 + 25)

7. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 7 Bruce Mayer, PE Chabot College Mathematics Example  Factor 5t 4 − 3125 C.Since t 2 − 25 is not prime, we continue factoring: 5(t 2 − 25)(t 2 + 25) = 5(t − 5)(t + 5)(t 2 + 25) SUM of squares with no common factor. It canNOT be factored!

8. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 8 Bruce Mayer, PE Chabot College Mathematics Example  Factor 5t 4 − 3125 D.Check: 5(t − 5)(t + 5)(t 2 + 25) = 5(t 2 − 25)(t 2 + 25) = 5(t 4 − 625) = 5t 4 − 3125  The factorization is VERIFIED as 5(t − 5)(t + 5)(t 2 + 25)

9. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 9 Bruce Mayer, PE Chabot College Mathematics Factor  2x 3 + 14x 2 + 3x + 21  SOLUTION A.We look for a common factor. There is none. B.Because there are four terms, try factoring by grouping: 2x 3 + 14x 2 + 3x + 21 = (2x 3 + 14x 2 ) + (3x + 21) = 2x 2 (x + 7) + 3 (x + 7) = (x + 7)(2x 2 + 3)

10. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 10 Bruce Mayer, PE Chabot College Mathematics Factor  2x 3 + 14x 2 + 3x + 21 C.Nothing can be factored further, so we have factored completely. D.Check by Forward FOIL: (x + 7)(2x 2 + 3) = 2x 3 + 3x + 14x 2 + 21 = 2x 3 + 14x 2 + 3x + 21 

11. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 11 Bruce Mayer, PE Chabot College Mathematics Factor  −x 5 − 2x 4 + 24x 3  SOLUTION A.We note that there is a common factor, −x 3 : −x 5 − 2x 4 + 24x 3 = −x 3 (x 2 + 2x − 24) B.The factor x 2 + 2x − 24 is not a perfect-square trinomial. We factor it by FOIL trial and error: −x 5 − 2x 4 + 34x 3 = −x 3 (x 2 + 2x − 24) = −x 3 (x − 4)(x + 6)

12. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 12 Bruce Mayer, PE Chabot College Mathematics Factor  −x 5 − 2x 4 + 24x 3 C.Nothing can be factored further, so we have factored completely D.Check: −x 3 (x − 4)(x + 6) = −x 3 (x 2 + 2x − 24) = −x 5 − 2x 4 + 24x 3 

13. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 13 Bruce Mayer, PE Chabot College Mathematics Factor  x 2 − 18x + 81  SOLUTION A.Look for a common factor. There is none. B.This polynomial is a perfect-square trinomial. Factor accordingly: x 2 − 18x + 81 = x 2 − 2  9  x + 9 2 = (x − 9)(x − 9) = (x − 9) 2

14. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 14 Bruce Mayer, PE Chabot College Mathematics Factor  x 2 − 18x + 81 C.Nothing can be factored further, so we have factored completely. D.Check: (x − 9)(x − 9) = x 2 − 18x + 81. 

15. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 15 Bruce Mayer, PE Chabot College Mathematics Factor  12x 2 y 3 + 20x 3 y 4 + 4x 2 y 5  SOLUTION A.We first factor out the largest common factor, 4x 2 y 3 : 4x 2 y 3 (3 + 5xy + y 2 ) B.The constant term in 3 + 5xy + y 2 is not a square, so we do not have a perfect- square trinomial. It cannot be factored using grouping or trial and error. The Trinomial term cannot be factored.

16. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 16 Bruce Mayer, PE Chabot College Mathematics Factor  12x 2 y 3 + 20x 3 y 4 + 4x 2 y 5 C.Nothing can be factored further, so we have factored completely D.Check: 4x 2 y 3 (3 + 5xy + y 2 ) = 12x 2 y 3 + 20x 3 y 4 + 4x 2 y 5 

17. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 17 Bruce Mayer, PE Chabot College Mathematics Factor  ab + ac + wb + wc  SOLUTION A. We look for a common factor. There is none. B. There are four terms. We try factoring by grouping: ab + ac + wb + wc = a(b + c) + w(b + c) = (b + c)(a + w)

18. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 18 Bruce Mayer, PE Chabot College Mathematics Factor  ab + ac + wb + wc C.Nothing can be factored further, so we have factored completely. D.Check by FOIL Multiplication: (b + c)(a + w) = ba + bw + ca + cw = ab + ac + wb + wc 

19. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 19 Bruce Mayer, PE Chabot College Mathematics Factor  36x 2 + 36xy + 9y 2  SOLUTION A.Look for common factor. The GCF is 9, but Let’s hold off for now B.There are three terms. Note that the first term and the last term are squares: 36x 2 = (6x) 2 and 9y 2 = (3y) 2.  We see that twice the product of 6x and 3y is the middle term, 2  6x  3y = 36xy, so the trinomial is a perfect square.

20. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 20 Bruce Mayer, PE Chabot College Mathematics Factor  36x 2 + 36xy + 9y 2 B.To Factor the Trinomial Square, we write the binomial squared: 36x 2 + 36xy + 9y 2 = (6x + 3y) 2 = (6x+3y)(6x+3y) = 3(2x + y)3(2x + y) = 3∙3(2x + y)(2x + y) = 9(2x + y) 2 C.Cannot Factor Further. D.Check: 9(2x + y) 2 = 9(2x + y)(2x + y) = 9(4x 2 + 2xy + 2yx + y 2 ) = 36x 2 + 36xy + 9y 2 

21. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 21 Bruce Mayer, PE Chabot College Mathematics Factor  a 8 − 16b 4  SOLUTION A.We look for a common factor. There is none. B.There are two terms. Since a 8 = (a 4 ) 2 and 16b 4 = (4b 2 ) 2, we see that we have a difference of squares  (a 4 ) 2 − (4b 2 ) 2 Thus, a 8 − 16b 4 = (a 4 + 4b 2 )(a 4 − 4b 2 )

22. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 22 Bruce Mayer, PE Chabot College Mathematics Factor  a 8 − 16b 4 C.The factor (a 4 − 4b 2 ) is itself a difference of squares. Thus, (a 4 − 4b 2 ) = (a 2 − 2b)(a 2 + 2b) D.Check: (a 4 + 4b 2 )(a 2 − 2b)(a 2 + 2b) = (a 4 + 4b 2 )(a 4 − 4b 2 ) = a 8 − 16b 4 

23. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example  Factor: 4x 2 – 14x + 12  SOLUTION  Look for a common factor  Find “2”: 4x 2 – 14x + 12 = 2(2x 2 – 7x + 6)  The other factor has three terms. The trinomial is not a square. Try to FOIL factor using trial and error 4x 2 – 14x + 12 = 2(2x – 3)(x – 2)  Cannot Factor Further; Check Later

24. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 24 Bruce Mayer, PE Chabot College Mathematics Example  18y 9 – 27y 8  SOLUTION  Look for a common factor – Find 9y 8 18y 9 – 27y 8 = 9y 8 (2y – 3)  The other factor has two terms but is not a difference of squares and not a sum or difference of cubes  No factor with more than one term can be factored further  Check: 9y 8 (2y − 3) = 18y 9 − 27y 8 

25. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 25 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §5.6 Exercise Set 28, 36, 62, 68, 78, 82, 86  Find & Factor the Trinomial Squares

26. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 26 Bruce Mayer, PE Chabot College Mathematics All Done for Today Factoring difference of 2 Squares

27. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 27 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix –

28. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 28 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

29. BMayer@ChabotCollege.edu MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 29 Bruce Mayer, PE Chabot College Mathematics