1. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §5.5 Factor Special Forms

2. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §5.4 → Factoring TriNomials  Any QUESTIONS About HomeWork §5.4 → HW-23 5.4 MTH 55

3. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 3 Bruce Mayer, PE Chabot College Mathematics §5.5 Factoring Special Forms  Factoring Perfect-Square Trinomials and Differences of Squares Recognizing Perfect-Square Trinomials Factoring Perfect-Square Trinomials Recognizing Differences of Squares Factoring Differences of Squares Factoring SUM of Two Cubes Facting DIFFERENCE of Two Cubes

4. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 4 Bruce Mayer, PE Chabot College Mathematics Recognizing Perfect-Sq Trinoms  A trinomial that is the square of a binomial is called a perfect-square trinomial A 2 + 2AB + B 2 = (A + B) 2 ; A 2 − 2AB + B 2 = (A − B) 2  Reading the right sides first, we see that these equations can be used to factor perfect-square trinomials.

5. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 5 Bruce Mayer, PE Chabot College Mathematics Recognizing Perfect-Sq Trinoms  Note that in order for the trinomial to be the square of a binomial, it must have the following: 1. Two terms, A 2 and B 2, must be squares, such as: 9, x 2, 100y 2, 25w 2 2. Neither A 2 or B 2 is being SUBTRACTED. 3. The remaining term is either 2  A  B or −2  A  B where A & B are the square roots of A 2 & B 2

6. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 6 Bruce Mayer, PE Chabot College Mathematics Example  Trinom Sqs  Determine whether each of the following is a perfect-square trinomial. a) x 2 + 8x + 16 b) t 2 − 9t − 36 c) 25x 2 + 4 – 20x  SOLUTION a) x 2 + 8x + 16 1.Two terms, x 2 and 16, are squares. 2.Neither x 2 or 16 is being subtracted. 3.The remaining term, 8x, is 2  x  4, where x and 4 are the square roots of x 2 and 16

7. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 7 Bruce Mayer, PE Chabot College Mathematics Example  Trinom Sqs  SOLUTION b) t 2 – 9t – 36 1.Two terms, t 2 and 36, are squares. But 2.But 36 is being subtracted so t 2 – 9t – 36 is not a perfect-square trinomial.  SOLUTION c) 25x 2 + 4 – 20x It helps to write it in descending order. 25x 2 – 20x + 4

8. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 8 Bruce Mayer, PE Chabot College Mathematics Example  Trinom Sqs  SOLUTION c) 25x 2 – 20x + 4 1.Two terms, 25x 2 and 4, are squares. 2.There is no minus sign before 25x 2 or 4. 3.Twice the product of the square roots is 2  5x  2, is 20x, the opposite of the remaining term, –20x  Thus 25x 2 – 20x + 4 is a perfect-square trinomial.

9. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 9 Bruce Mayer, PE Chabot College Mathematics Factoring a Perfect-Square Trinomial  The Two Type of Perfect-Squares A 2 + 2AB + B 2 = (A + B) 2 A 2 − 2AB + B 2 = (A − B) 2

10. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 10 Bruce Mayer, PE Chabot College Mathematics Example  Factor Perf. Sqs  Factor: a) x 2 + 8x + 16 b) 25x 2 − 20x + 4  SOLUTION a) x 2 + 8x + 16 = x 2 + 2  x  4 + 4 2 = (x + 4) 2 A 2 + 2 A B + B 2 = (A + B) 2

11. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example  Factor Perf. Sqs  Factor: a) x 2 + 8x + 16 b) 25x 2 − 20x + 4  SOLUTION b) 25x 2 – 20x + 4 = (5x) 2 – 2  5x  2 + 2 2 = (5x – 2) 2 A 2 – 2 A B + B 2 = (A – B) 2

12. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 12 Bruce Mayer, PE Chabot College Mathematics Example  Factor 16a 2 – 24ab + 9b 2  SOLUTION 16a 2 − 24ab + 9b 2 = (4a) 2 − 2(4a)(3b) + (3b) 2 = (4a − 3b) 2  CHECK: (4a − 3b)(4a − 3b) = 16a 2 − 24ab + 9b 2   The factorization is (4a − 3b) 2.

13. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 13 Bruce Mayer, PE Chabot College Mathematics Expl  Factor 12a 3 – 108a 2 + 243a  SOLUTION  Always look for a common factor. This time there is one. Factor out 3a. 12a 3 − 108a 2 + 243a = 3a(4a 2 − 36a + 81) = 3a[(2a) 2 − 2(2a)(9) + 9 2 ] = 3a(2a − 9) 2  The factorization is 3a(2a − 9) 2

14. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 14 Bruce Mayer, PE Chabot College Mathematics Recognizing Differences of Squares  An expression, like 25x 2 − 36, that can be written in the form A 2 − B 2 is called a difference of squares.  Note that for a binomial to be a difference of squares, it must have the following. 1.There must be two expressions, both squares, such as: 9, x 2, 100y 2, 36y 8 2.The terms in the binomial must have different signs.

15. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 15 Bruce Mayer, PE Chabot College Mathematics Difference of 2-Squares  Diff of 2 Sqs → A 2 − B 2  Note that in order for a term to be a square, its coefficient must be a perfect square and the power(s) of the variable(s) must be even. For Example 25x 4 − 36 – 25 = 5 2 – The Power on x is even at 4 → x 4 = (x 2 ) 2 – Also, in this case 36 = 6 2

16. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 16 Bruce Mayer, PE Chabot College Mathematics Example  Test Diff of 2Sqs  Determine whether each of the following is a difference of squares. a) 16x 2 − 25b) 36 − y 5 c) −x 12 + 49  SOLUTION a) 16x 2 − 25 1.The 1st expression is a sq: 16x 2 = (4x) 2 The 2nd expression is a sq: 25 = 5 2 2.The terms have different signs.  Thus, 16x 2 − 25 is a difference of squares, (4x) 2 − 5 2

17. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 17 Bruce Mayer, PE Chabot College Mathematics Example  Test Diff of 2Sqs  SOLUTION b) 36 − y 5 1.The expression y 5 is not a square.  Thus, 36 − y 5 is not a diff of squares  SOLUTION c) −x 12 + 49 1.The expressions x 12 and 49 are squares: x 12 = (x 6 ) 2 and 49 = 7 2 2.The terms have different signs.  Thus, −x 12 + 49 is a diff of sqs, 7 2 − (x 6 ) 2

18. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 18 Bruce Mayer, PE Chabot College Mathematics Factoring Diff of 2 Squares  A 2 − B 2 = (A + B)(A − B)  The Gray Area by Square Subtraction  The Gray Area by (LENGTH)(WIDTH)

19. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example  Factor Diff of Sqs  Factor: a) x 2 − 9b) y 2 − 16w 2  SOLUTION a)x 2 − 9 = x 2 – 3 2 = (x + 3)(x − 3) A 2 − B 2 = (A + B)(A − B) b) y 2 − 16w 2 = y 2 − (4w) 2 = (y + 4w)(y − 4w) A 2 − B 2 = (A + B) (A − B)

20. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example  Factor Diff of Sqs  Factor: c) 25 − 36a 12 d) 98x 2 − 8x 8  SOLUTION c) 25 − 36a 12 = 5 2 − (6a 6 ) 2 = (5 + 6a 6 )(5 − 6a 6 ) d) 98x 2 − 8x 8 Always look for a common factor. This time there is one, 2x 2 : 98x 2 − 8x 8 = 2x 2 (49 − 4x 6 ) = 2x 2 [(7 2 − (2x 3 ) 2 ] = 2x 2 (7 + 2x 3 )(7 − 2x 3 )

21. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 21 Bruce Mayer, PE Chabot College Mathematics Grouping to Expose Diff of Sqs  Sometimes a Clever Grouping will reveal a Perfect-Sq TriNomial next to another Squared Term  Example  Factor m 2 − 4b 4 + 14m + 49  rearranging  m 2 + 14m + 49 − 4b 4  GROUPING  (m 2 + 14m + 49) − 4b 4

22. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 22 Bruce Mayer, PE Chabot College Mathematics Grouping to Expose Diff of Sqs  Example  Factor m 2 − 4b 4 + 14m + 49 Recognize m 2 + 14m + 49 as Perfect Square Trinomial → (m+7) 2 Also Recognize 4b 4 as a Sq → (2b) 2 (m 2 + 14m + 49) − 4b 4  Perfect Sqs  (m + 7) 2 − (2b 2 ) 2  In Diff-of-Sqs Formula: A→m+7; B→2b 2

23. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 23 Bruce Mayer, PE Chabot College Mathematics Grouping to Expose Diff of Sqs  Example  Factor m 2 − 4b 4 + 14m + 49 (m + 7) 2 − (2b 2 ) 2  Diff-of-Sqs → (A − B)(A + B)  ([m+7] − 2b 2 )([m + 7] + 2b 2 )  Simplify → ReArrange  (−2b 2 + m + 7)(2b 2 + m + 7)  The Check is Left for us to do Later

24. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 24 Bruce Mayer, PE Chabot College Mathematics Factoring Two Cubes  The principle of patterns applies to the sum and difference of two CUBES. Those patterns SUM of Cubes DIFFERENCE of Cubes

25. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 25 Bruce Mayer, PE Chabot College Mathematics TwoCubes SIGN Significance  Carefully note the Sum/Diff of Two-Cubes Sign Pattern SAME SignOPP Sign SAME SignOPP Sign

26. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 26 Bruce Mayer, PE Chabot College Mathematics Example: Factor x 3 + 64  Factor Recognize Pattern as Sum of CUBES Determine Values that were CUBED Map Values to Formula Substitute into Formula Simplify and CleanUp

27. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 27 Bruce Mayer, PE Chabot College Mathematics Example: Factor 8w 3 −27z 3  Factor Recognize Pattern as Difference of CUBES Determine CUBED Values Map Values to Formula Sub into Formula Simplify & CleanUp Simplify by Properties of Exponents

28. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 28 Bruce Mayer, PE Chabot College Mathematics Example: Check 8w 3 −27z 3  Check Use Comm & Assoc. properties, and Adding-to-Zero Use Distributive property 

29. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 29 Bruce Mayer, PE Chabot College Mathematics Sum & Difference Summary  Difference of Two SQUARES  SUM of Two CUBES  Difference of Two CUBES

30. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 30 Bruce Mayer, PE Chabot College Mathematics Factoring Completely  Sometimes, a complete factorization requires two or more steps. Factoring is complete when no factor can be factored further.  Example: Factor 5x 4 − 3125 May have the Difference-of-2sqs TWICE

31. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 31 Bruce Mayer, PE Chabot College Mathematics Factoring Completely  SOLUTION 5x 4 − 3125 = 5(x 4 − 625) = 5[(x 2 ) 2 − 25 2 ] = 5(x 2 − 25)(x 2 + 25) = 5(x − 5)(x + 5)(x 2 + 25)  The factorization: 5(x − 5)(x + 5)(x 2 + 25)

32. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 32 Bruce Mayer, PE Chabot College Mathematics Factoring Tips 1. Always look first for a common factor. If there is one, factor it out. 2. Be alert for perfect-square trinomials and for binomials that are differences of squares. Once recognized, they can be factored without trial and error. 3. Always factor completely. 4. Check by multiplying.

33. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 33 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §5.5 Exercise Set 14, 22, 48, 74, 94, 110  The SUM (Σ) & DIFFERENCE (Δ) of Two Cubes

34. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 34 Bruce Mayer, PE Chabot College Mathematics All Done for Today Sum of Two Cubes

35. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 35 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix –

36. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 36 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

37. BMayer@ChabotCollege.edu MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 37 Bruce Mayer, PE Chabot College Mathematics