W A T K I N S - J O H N S O N C O M P A N Y Semiconductor Equipment Group

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Chabot Mathematics
5.6 FactoringStrategies
Bruce Mayer, PE Licensed Electrical Mechanical
EngineerBMayer_at_ChabotCollege.edu
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Review

• Any QUESTIONS About
• 5.5 ? Factoring TriNomials, Special Forms
• Any QUESTIONS About HomeWork
• 5.5 ? HW-19

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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)

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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.

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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

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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)

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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!
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Example ? Factor 5t4 - 3125

• Check5(t - 5)(t 5)(t2 25)
• 5(t2 - 25)(t2 25)
• 5(t4 - 625)
• 5t4 - 3125
• The factorization is VERIFIED as 5(t - 5)(t
  5)(t2 25)

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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)

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Factor ? 2x3 14x2 3x 21

• Nothing can be factored further, so we have
  factored completely.
• Check by Forward FOIL
• (x 7)(2x2 3) 2x3 3x 14x2 21
• 2x3 14x2 3x 21

?
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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)

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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

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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

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Factor ? x2 - 18x 81

• Nothing can be factored further, so we have
  factored completely.
• Check
• (x - 9)(x - 9) x2 - 18x 81.

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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.

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Factor ? 12x2y3 20x3y4 4x2y5

• Nothing can be factored further, so we have
  factored completely
• Check
• 4x2y3(3 5xy y2)
• 12x2y3 20x3y4 4x2y5

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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)

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Factor ? ab ac wb wc

• Nothing can be factored further, so we have
  factored completely.
• Check by FOIL Multiplication
• (b c)(a w) ba bw ca cw
• ab ac wb wc

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Factor ? 36x2 36xy 9y2

• SOLUTION
• Look for common factor. The GCF is 9, but Lets
  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.

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Factor ? 36x2 36xy 9y2

• To Factor the Trinomial Square, we write the
  binomial squared
• 36x2 36xy 9y2 (6x 3y)2
• (6x3y)(6x3y) 3(2x y)3(2x y)
• 33(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

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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)

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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

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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

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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 ?

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WhiteBoard Work

• Problems From 5.6 Exercise Set
• 28, 36, 62, 68, 78, 82, 86

• Find Factorthe TrinomialSquares

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All Done for Today
Factoringdifference of 2 Squares
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Chabot Mathematics
Appendix
Bruce Mayer, PE Licensed Electrical Mechanical
EngineerBMayer_at_ChabotCollege.edu

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Graph y x

• Make T-table

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