Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

1. Chabot Mathematics §5.5 FactorSpecial Forms Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. MTH 55 5.4 Review § • Any QUESTIONS About • §5.4 → Factoring TriNomials • Any QUESTIONS About HomeWork • §5.4 → HW-14

3. §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. Recognizing Perfect-Sq Trinoms • A trinomial that is the square of a binomial is called a perfect-square trinomial A2 + 2AB + B2 = (A + B)2 A2− 2AB + B2 = (A−B)2 • Reading the right sides first, we see that these equations can be used to factor perfect-square trinomials. • A2 + 2AB + B2 = (A + B)(A + B) • A2− 2AB + B2 = (A−B)(A−B)

5. 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, A2 and B2, must be squares, such as: 9, x2, 100y2, 25w2 2. Neither A2 or B2 is being SUBTRACTED. 3. The remaining term is either 2  A  B or −2  A  B • where A &B are the square roots of A2 & B2

6. Example  Trinom Sqs • Determine whether each of the following is a perfect-square trinomial. a) x2 + 8x + 16 b) t2− 9t− 36 c) 25x2 + 4 – 20x • SOLUTION a) x2 + 8x + 16 • Two terms, x2 and 16, are squares. • Neither x2 or 16 is being subtracted. • The remaining term, 8x, is 2x4, where x and 4 are the square roots of x2 and 16

7. Example  Trinom Sqs • SOLUTION b) t2– 9t– 36 • Two terms, t2 and 36, are squares. But • But 36 is being subtracted so t2– 9t– 36 is nota perfect-square trinomial. • SOLUTION c) 25x2 + 4 – 20x It helps to write it in descending order. 25x2– 20x + 4

8. Example  Trinom Sqs • SOLUTION c) 25x2− 20x + 4 • Two terms, 25x2 and 4, are squares. • There is no minus sign before 25x2 or 4. • Twice the product of the square roots is 2  5x 2, is 20x, the opposite of the remaining term, −20x • Thus 25x2− 20x + 4 is a perfect-square trinomial.

9. Factoring a Perfect-Square Trinomial • The Two Types of Perfect-Squares A2 + 2AB + B2 = (A + B)2 A2− 2AB + B2 = (A−B)2

10. Example  Factor Perf. Sqs • Factor: a) x2 + 8x + 16 b) 25x2− 20x + 4 • SOLUTION a) x2 + 8x + 16 = x2 + 2  x  4 + 42 = (x + 4)2 A2 + 2 A B + B2 = (A + B)2

11. Example  Factor Perf. Sqs • Factor: a) x2 + 8x + 16 b) 25x2− 20x + 4 • SOLUTION b) 25x2– 20x + 4 = (5x)2–2  5x  2 + 22 = (5x– 2)2 A2– 2 A B + B2 = (A – B)2

12. Example  Factor 16a2– 24ab + 9b2 • SOLUTION 16a2− 24ab + 9b2= (4a)2− 2(4a)(3b) + (3b)2 = (4a− 3b)2 = (4a− 3b)(4a− 3b) • CHECK: (4a− 3b)(4a− 3b) = 16a2− 24ab + 9b2  • The factorization is (4a− 3b)2.

13. Expl  Factor 12a3 –108a2 + 243a • SOLUTION • Always look for a common factor. This time there is one. Factor out 3a. 12a3− 108a2 + 243a = 3a(4a2− 36a + 81) = 3a[(2a)2− 2(2a)(9) + 92] = 3a(2a− 9)2 • The factorization is 3a(2a− 9)2

14. Recognizing Differences of Squares • An expression, like 25x2− 36, that can be written in the form A2−B2 is called a difference of squares. • Note that for a binomial to be a difference of squares, it must have the following. • There must be two expressions, both squares, such as: 9, x2, 100y2, 36y8 • The terms in the binomial must have different signs.

15. Difference of 2-Squares • Diff of 2 Sqs → A2−B2 • 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 25x4− 36 • 25 = 52 • The Power on x is even at 4 → x4 = (x2)2 • Also, in this case 36 = 62

16. Example  Test Diff of 2Sqs • Determine whether each of the following is a difference of squares. a) 16x2− 25 b) 36 −y5 c) −x12 + 49 • SOLUTION a) 16x2− 25 • The 1st expression is a sq: 16x2 = (4x)2 The 2nd expression is a sq: 25 = 52 • The terms have different signs. • Thus, 16x2− 25 is a difference of squares, (4x)2− 52

17. Example  Test Diff of 2Sqs • SOLUTION b) 36 −y5 • The expression y5 is not a square. • Thus, 36 −y5 is not a diff of squares • SOLUTION c) −x12 + 49 • The expressions x12 and 49 are squares:x12 = (x6)2 and 49 = 72 • The terms have different signs. • Thus, −x12 + 49 is a diff of sqs, 72− (x6)2

18. Factoring Diff of 2 Squares • A2−B2 = (A + B)(A−B) • The Gray Area by Square Subtraction • The Gray Area by(LENGTH)(WIDTH)

19. Example  Factor Diff of Sqs • Factor: a) x2− 9 b) y2− 16w2 • SOLUTION a) x2− 9 = x2– 32 = (x + 3)(x− 3) A2−B2 = (A + B)(A−B) b) y2− 16w2 = y2− (4w)2 = (y + 4w)(y− 4w) A2−B2 = (A + B) (A−B)

20. Example  Factor Diff of Sqs • Factor: c) 25 − 36a12 d) 98x2− 8x8 • SOLUTION c) 25 − 36a12 = 52− (6a6)2 = (5 + 6a6)(5 − 6a6) d) 98x2− 8x8 Alwayslook for a common factor. This time there is one, 2x2: 98x2− 8x8 = 2x2(49 − 4x6) = 2x2[(72− (2x3)2] = 2x2(7 + 2x3)(7 − 2x3)

21. Grouping to Expose Diff of Sqs • Sometimes a Clever Grouping will reveal a Perfect-Sq TriNomial next to another Squared Term • Example Factor m2−4b4 + 14m + 49  rearranging  m2 + 14m + 49 − 4b4  GROUPING  (m2 + 14m + 49) − 4b4

22. Grouping to Expose Diff of Sqs • Example Factor m2− 4b4 + 14m + 49 • Recognize m2 + 14m + 49 as Perfect Square Trinomial → (m+7)2 • Also Recognize 4b4 as a Sq → (2b)2 (m2 + 14m + 49) − 4b4  Perfect Sqs  (m + 7)2− (2b2)2 • In Diff-of-Sqs Formula: A→m+7; B→2b2

23. Grouping to Expose Diff of Sqs • Example Factor m2− 4b4 + 14m + 49 (m + 7)2− (2b2)2  Diff-of-Sqs → (A − B)(A + B)  ([m+7] − 2b2)([m + 7] + 2b2)  Simplify → ReArrange  (−2b2 + m + 7)(2b2 + m + 7) • The Check is Left for us to do Later

24. 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. TwoCubes SIGN Significance • Carefully note the Sum/Diff of Two-Cubes Sign Pattern SAME Sign OPP Sign SAME Sign OPP Sign

26. Example: Factor x3 + 64 • Factor Recognize Pattern as Sum of CUBES Determine Values that were CUBED Map Values to Formula Substitute into Formula Simplify and CleanUp

27. Example: Factor 8w3−27z3 • Factor Recognize Pattern as Difference of CUBES Determine CUBED Values Simplify by Properties of Exponents Map Values to Formula Sub into Formula Simplify & CleanUp

28. Example: Check 8w3−27z3 • Check Use Distributive property Use Comm & Assoc. properties, and Adding-to-Zero 

29. Sum & Difference Summary • Difference of Two SQUARES • SUM of Two CUBES • Difference of Two CUBES

30. Factoring Completely • Sometimes, a complete factorization requires two or more steps. Factoring is complete when no factor can be factored further. • Example: Factor 5x4− 3125 • May have the Difference-of-2sqs TWICE

31. Factoring Completely • SOLUTION 5x4− 3125 = 5(x4− 625) = 5[(x2)2− 252] = 5(x2− 25)(x2 + 25) = 5(x− 5)(x + 5)(x2 + 25) • The factorization: 5(x− 5)(x + 5)(x2 + 25)

32. Factoring Tips • Always look first for a common factor. If there is one, factor it out. • Be alert for perfect-square trinomials and for binomials that are differences of squares. • Once recognized, they can be factored without trial and error. • Always factor completely. • Check by multiplying.

33. WhiteBoard Work • Problems From §5.5 Exercise Set • 14, 22, 48, 74, 94, 110 • The SUM (Σ) & DIFFERENCE (Δ) of Two Cubes

34. All Done for Today Sum ofTwoCubes

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

36. Graph y = |x| • Make T-table