§5.6 Factoring Strategies Chabot Mathematics §5.6 Factoring Strategies Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

5.5 Review § Any QUESTIONS About Any QUESTIONS About HomeWork MTH 55 Review § Any QUESTIONS About §5.[4 &5] → Factoring: TriNomials, Special Forms Any QUESTIONS About HomeWork §5.[4 & 5] → HW-[14 & 15]

To Factor a Polynomial 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. Then look at the number of terms TWO Terms: If you have a Difference of SQUARES, factor accordingly: A2 − B2 = (A − B)(A + B)

To Factor a Polynomial TWO Terms: If you have a SUM of CUBES, factor accordingly: A3 + B3 = (A + B)(A2 − AB + B2) TWO Terms: If you have a DIFFERENCE of CUBES, factor accordingly: A3 − B3 = (A − B)(A2 + AB + B2) THREE Terms: If the trinomial is a perfect-square trinomial, factor accordingly: A2 + 2AB + B2 = (A + B)2 or A2 – 2AB + B2 = (A – B)2.

To Factor a Polynomial THREE Terms: If it is not a perfect-square trinomial, try using FOIL Guessing FOUR Terms: Try factoring by grouping Always factor completely: When a factor can itself be factored, be sure to factor it. Remember that some polynomials, like A2 + B2, are PRIME CHECK by Multiplying the Factors

Choosing the Right Method Example: Factor 5t4 − 3125 SOLUTION Look for a common factor: 5t4 − 3125 = 5(t4 − 625). The factor t4 − 625 is a diff of squares: (t2)2 − 252. We factor it, being careful to rewrite the 5 from step (A): 5(t4 − 625) = 5(t2 − 25)(t2 + 25)

SUM of squares with no common factor. It canNOT be factored! Example  Factor 5t4 − 3125 Since t2 − 25 is not prime, we continue factoring: 5(t2 − 25)(t2 + 25) = 5(t − 5)(t + 5)(t2 + 25) SUM of squares with no common factor. It canNOT be factored!

Example  Factor 5t4 − 3125 Check: 5(t − 5)(t + 5)(t2 + 25) The factorization is 5(t − 5)(t + 5)(t2 + 25)

Factor  2x3 + 14x2 + 3x + 21 SOLUTION We look for a common factor. There is none. Because there are four terms, try factoring by grouping: 2x3 + 14x2 + 3x + 21 = (2x3 + 14x2) + (3x + 21) = 2x2(x + 7) + 3(x + 7) = (x + 7)(2x2 + 3)

Factor  2x3 + 14x2 + 3x + 21 Nothing can be factored further, so we have factored completely. Check: (x + 7)(2x2 + 3) = 2x3 + 3x + 14x2 + 21 = 2x3 + 14x2 + 3x + 21 

Factor  −x5 − 2x4 + 24x3 SOLUTION We note that there is a common factor, −x3: −x5 − 2x4 + 24x3 = −x3(x2 + 2x − 24) The factor x2 + 2x − 24 is not a perfect-square trinomial. We factor it by FOIL trial and error: −x5 − 2x4 + 34x3 = −x3(x2 + 2x − 24) = −x3(x − 4)(x + 6)

Factor  −x5 − 2x4 + 24x3 Nothing can be factored further, so we have factored completely Check: −x3(x − 4)(x + 6) = −x3(x2 + 2x − 24) = −x5 − 2x4 + 24x3 

Factor  x2 − 18x + 81 SOLUTION Look for a common factor. There is none. This polynomial is a perfect-square trinomial. Factor accordingly: x2 − 18x + 81 = x2 − 2  9  x + 92 = (x − 9)(x − 9) = (x − 9)2

Factor  x2 − 18x + 81 Nothing can be factored further, so we have factored completely. Check: (x − 9)(x − 9) = x2 − 18x + 81. 

Factor  12x2y3 + 20x3y4 + 4x2y5 SOLUTION We first factor out the largest common factor, 4x2y3: 4x2y3(3 + 5xy + y2) The constant term in 3 + 5xy + y2 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.

Factor  12x2y3 + 20x3y4 + 4x2y5 Nothing can be factored further, so we have factored completely Check: 4x2y3(3 + 5xy + y2) = 12x2y3 + 20x3y4 + 4x2y5 

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)

Factor  ab + ac + wb + wc Nothing can be factored further, so we have factored completely. Check: (b + c)(a + w) = ba + bw + ca + cw = ab + ac + wb + wc 

Factor  36x2 + 36xy + 9y2 SOLUTION Look for common factor. The GCF is 9, but Let’s hold off for now There are three terms. Note that the first term and the last term are squares: 36x2 = (6x)2 and 9y2 = (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.

Factor  36x2 + 36xy + 9y2 To Factor the Trinomial Square, we write the binomial squared: 36x2 + 36xy + 9y2 = (6x + 3y)2 = (6x+3y)(6x+3y) = 3(2x + y)3(2x + y) = 3∙3(2x + y)(2x + y) = 9(2x + y)2 Cannot Factor Further. Check: 9(2x + y)2 = 9(2x + y)(2x + y) = 9(4x2 + 2xy + 2yx + y2) = 36x2 + 36xy + 9y2 

Factor  a8 − 16b4 SOLUTION We look for a common factor. There is none. There are two terms. Since a8 = (a4)2 and 16b4 = (4b2)2, we see that we have a difference of squares  (a4)2 − (4b2)2 Thus, a8 − 16b4 = (a4 + 4b2)(a4 − 4b2)

Factor  a8 − 16b4 The factor (a4 − 4b2) is itself a difference of squares. Thus, (a4 − 4b2) = (a2 − 2b)(a2 + 2b) Check: (a4 + 4b2)(a2 − 2b)(a2 + 2b) = (a4 + 4b2)(a4 − 4b2) = a8 − 16b4 

Example  Factor: 4x2 – 14x + 12 SOLUTION Look for a common factor  Find “2”: 4x2 – 14x + 12 = 2(2x2 – 7x + 6) The other factor has three terms. The trinomial is not a square. Try to FOIL factor using trial and error 4x2 – 14x + 12 = 2(2x – 3)(x – 2) Cannot Factor Further; Check Later

Example  18y9 – 27y8 SOLUTION Look for a common factor – Find 9y8 18y9 – 27y8 = 9y8(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: 9y8(2y − 3) = 18y9 − 27y8 

WhiteBoard Work Problems From §5.6 Exercise Set 28, 36, 62, 68, 78, 82, 86 Find & Factor the Trinomial Squares

Factoring difference of 2 Squares All Done for Today Factoring difference of 2 Squares

Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu –

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