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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] - PowerPoint PPT Presentation

Chabot Mathematics. §5.5 Factor Special Forms. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] MTH 55. 5.4. Review §. Any QUESTIONS About §5.4 → Factoring TriNomials Any QUESTIONS About HomeWork §5.4 → HW-14. §5.5 Factoring Special Forms.

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§5.5 FactorSpecial Forms

Bruce Mayer, PE

Licensed Electrical & Mechanical [email protected]

5.4

Review §

• §5.4 → Factoring TriNomials

• §5.4 → HW-14

• 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

• 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)

• 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

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

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

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.

• The Two Types of Perfect-Squares

A2 + 2AB + B2 = (A + B)2

A2− 2AB + B2 = (A−B)2

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

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

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.

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

• 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.

• 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

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

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

• A2−B2 = (A + B)(A−B)

• The Gray Area by Square Subtraction

• The Gray Area by(LENGTH)(WIDTH)

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)

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)

• 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

• 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

• 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

• The principle of patterns applies to the sum and difference of two CUBES. Those patterns

• SUM of Cubes

• DIFFERENCE of Cubes

• Carefully note the Sum/Diff of Two-Cubes Sign Pattern

SAME Sign

OPP Sign

SAME Sign

OPP Sign

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

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

Example: Check 8w3−27z3

• Check

Use Distributive property

Use Comm & Assoc. properties, and Adding-to-Zero

• Difference of Two SQUARES

• SUM of Two CUBES

• Difference of Two CUBES

• 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

• 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)

• 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.

• Problems From §5.5 Exercise Set

• 14, 22, 48, 74, 94, 110

• The SUM (Σ) & DIFFERENCE (Δ) of Two Cubes

Sum ofTwoCubes

Appendix

Bruce Mayer, PE

Licensed Electrical & Mechanical [email protected]

Graph y = |x|

• Make T-table