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

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 MTH 55

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)

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.

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

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 − We factor it, being careful to rewrite the 5 from step (A): 5(t 4 − 625) = 5(t 2 − 25)(t )

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 ) = 5(t − 5)(t + 5)(t ) SUM of squares with no common factor. It canNOT be factored!

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 ) = 5(t 2 − 25)(t ) = 5(t 4 − 625) = 5t 4 − 3125  The factorization is VERIFIED as 5(t − 5)(t + 5)(t )

MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 9 Bruce Mayer, PE Chabot College Mathematics Factor  2x x 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 x 2 + 3x + 21 = (2x x 2 ) + (3x + 21) = 2x 2 (x + 7) + 3 (x + 7) = (x + 7)(2x 2 + 3)

MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 10 Bruce Mayer, PE Chabot College Mathematics Factor  2x x 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 = 2x x 2 + 3x + 21 

MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 11 Bruce Mayer, PE Chabot College Mathematics Factor  −x 5 − 2x x 3  SOLUTION A.We note that there is a common factor, −x 3 : −x 5 − 2x x 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 x 3 = −x 3 (x 2 + 2x − 24) = −x 3 (x − 4)(x + 6)

MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 12 Bruce Mayer, PE Chabot College Mathematics Factor  −x 5 − 2x x 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 x 3 

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 = (x − 9)(x − 9) = (x − 9) 2

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 

MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 15 Bruce Mayer, PE Chabot College Mathematics Factor  12x 2 y x 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.

MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 16 Bruce Mayer, PE Chabot College Mathematics Factor  12x 2 y x 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 x 3 y 4 + 4x 2 y 5 

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)

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 

MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 19 Bruce Mayer, PE Chabot College Mathematics Factor  36x xy + 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.

MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt 20 Bruce Mayer, PE Chabot College Mathematics Factor  36x xy + 9y 2 B.To Factor the Trinomial Square, we write the binomial squared: 36x xy + 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 xy + 9y 2 

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 )

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 

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

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 

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

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

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

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

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