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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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Chabot Mathematics

§5.5 FactorSpecial Forms

Bruce Mayer, PE

Licensed Electrical & Mechanical [email protected]


Review

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
§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


Recognizing perfect sq trinoms
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)


Recognizing perfect sq trinoms1
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


Example trinom sqs
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 sqs1
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 sqs2
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.


Factoring a perfect square trinomial
Factoring 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
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 sqs1
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 16 a 2 24 ab 9 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 12 a 3 108 a 2 243 a
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


Recognizing differences of squares
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.


Difference of 2 squares
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


Example test diff of 2sqs
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 2sqs1
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


Factoring diff of 2 squares
Factoring Diff of 2 Squares

  • A2−B2 = (A + B)(A−B)

  • The Gray Area by Square Subtraction

  • The Gray Area by(LENGTH)(WIDTH)


Example factor diff of sqs
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 sqs1
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)


Grouping to expose diff of sqs
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


Grouping to expose diff of sqs1
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


Grouping to expose diff of sqs2
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


Factoring two cubes
Factoring Two Cubes

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

    • SUM of Cubes

  • DIFFERENCE of Cubes


Twocubes sign significance
TwoCubes SIGN Significance

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

SAME Sign

OPP Sign

SAME Sign

OPP Sign


Example factor x 3 64
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 8 w 3 27 z 3
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 8 w 3 27 z 3
Example: Check 8w3−27z3

  • Check

Use Distributive property

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


Sum difference summary
Sum & Difference Summary

  • Difference of Two SQUARES

  • SUM of Two CUBES

  • Difference of Two CUBES


Factoring completely
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


Factoring completely1
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)


Factoring tips
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.


Whiteboard work
WhiteBoard Work

  • Problems From §5.5 Exercise Set

    • 14, 22, 48, 74, 94, 110

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


All done for today
All Done for Today

Sum ofTwoCubes


Chabot Mathematics

Appendix

Bruce Mayer, PE

Licensed Electrical & Mechanical [email protected]


Graph y x
Graph y = |x|

  • Make T-table


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