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64 Solving Polynomial Equations Day 1 1 January 13, 2009 Dec 2111:45 PM 6-4 Solving Polynomial Equations Objective: Solve cubic polynomial equations by factoring.

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Page 1: 6-4 Solving Polynomial Equations Day 1staffweb.psdschools.org/kemotich/Mrs_Motichka/Algebra 2/Chapter … · 64 Solving Polynomial Equations Day 1 2 January 13, 2009 Dec 2111:46 PM

6­4 Solving Polynomial Equations Day 1 

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January 13, 2009

Dec 21­11:45 PM

6-4 Solving Polynomial Equations

Objective:Solve cubic polynomial equations by factoring.

Page 2: 6-4 Solving Polynomial Equations Day 1staffweb.psdschools.org/kemotich/Mrs_Motichka/Algebra 2/Chapter … · 64 Solving Polynomial Equations Day 1 2 January 13, 2009 Dec 2111:46 PM

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Factor each expression.

1. x2 - 2x - 15 2. x2 - 9x + 14 3. x2 + 6x + 5

Page 3: 6-4 Solving Polynomial Equations Day 1staffweb.psdschools.org/kemotich/Mrs_Motichka/Algebra 2/Chapter … · 64 Solving Polynomial Equations Day 1 2 January 13, 2009 Dec 2111:46 PM

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Think back to Ch 5...a quadratic difference of squares has a special factoring pattern.

Think forward to Ch 6...a cubic sum of cubes and difference of cubes also have special factoring patterns.

Page 4: 6-4 Solving Polynomial Equations Day 1staffweb.psdschools.org/kemotich/Mrs_Motichka/Algebra 2/Chapter … · 64 Solving Polynomial Equations Day 1 2 January 13, 2009 Dec 2111:46 PM

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Don't believe the pattern?

Multiply to verify...

(a + b)(a2 - ab + b2) = a(a2 - ab + b2) + b(a2 - ab + b2)

= a3 - a2b + ab2 + a2b - ab2 + b3

= a3 + b3

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Factoring a Sum or Difference of Cubes

Example #1: Factor x3 - 8 Example #2: Factor 8x3 + 1

x3 - 8 8x3 + 1

(x)3 - (2)3 (2x)3 + (1)3

(x - 2)(x2 + 2x + (2)2) (2x + 1)((2x)2 - 2x + (1)2)

(x - 2)(x2 + 2x + 4) (2x + 1)(4x2 - 2x + 1)

Page 6: 6-4 Solving Polynomial Equations Day 1staffweb.psdschools.org/kemotich/Mrs_Motichka/Algebra 2/Chapter … · 64 Solving Polynomial Equations Day 1 2 January 13, 2009 Dec 2111:46 PM

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Example #3: Solve 27x3 + 1 = 0. Find all complex roots.

27x3 + 1 = (3x)3 + (1)3

= (3x + 1)((3x)2 - 3x + (1)2)

= (3x + 1)(9x2 - 3x + 1)

Since 3x + 1 is a factor, x = -1/3 is a root.

Since 9x2 - 3x + 1 is a factor, we must use the quadratic formula to solve...

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January 13, 2009

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Page 8: 6-4 Solving Polynomial Equations Day 1staffweb.psdschools.org/kemotich/Mrs_Motichka/Algebra 2/Chapter … · 64 Solving Polynomial Equations Day 1 2 January 13, 2009 Dec 2111:46 PM

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Solve each Equation:

Example #4: x3 + 8 = 0 Example #5: 27x3 - 1 = 0

x3 + 8 = (x)3 + (2)3 27x3 - 1 = (3x)3 - (1)3

= (x + 2)(x2 - 2x + (2)2) = (3x - 1)((3x)2 + 3x + (1)2)

=(x + 2)(x2 - 2x + 4) =(3x - 1)(9x2 + 3x + 1)

x = -2 and x = ? x = 1/3 and x = ?

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January 13, 2009

Jan 13­8:28 AM

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Dec 22­1:54 AM

Homework:page 330 (12 - 17)

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