6­4 Solving Polynomial Equations Day 2

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January 14, 2009

Jan 6­6:59 AM

6­4 Day 2 Solving Polynomial Equations

Objective:Solve quartic polynomial equations by factoring.

6­4 Solving Polynomial Equations Day 2

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January 14, 2009

Jan 6­6:59 AM

Factoring by Using a Quadratic Pattern:

Example #1:  x4 ­ 2x2 ­ 8

Step 1:  Write in the pattern of a quadratic by substituting valuesx4 ­ 2x2 ­ 8 = (x2)2 ­ 2(x2) ­ 8    substitute a for x2

     = a2 ­ 2a ­ 8

Step 2:  Factor a2 ­ 2a ­ 8

a2 ­ 2a ­ 8 = (a ­ 4)(a + 2)

6­4 Solving Polynomial Equations Day 2

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January 14, 2009

Jan 6­6:59 AM

Example #1(continued...):  x4 ­ 2x2 ­ 8

Step 3:  Substitute back to the original variable and factor.

(a ­ 4)(a + 2) = (x2 ­ 4)(x2 + 2)    

         = (x ­ 2)(x + 2)(x2 + 2) 

Answer:       The factored form of x4 ­ 2x2 ­ 8 is (x ­ 2)(x + 2)(x2 + 2).

6­4 Solving Polynomial Equations Day 2

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January 14, 2009

Jan 6­7:00 AM

Factor each expression:

Example #2: x4 + 7x2 + 6 Example #3: x4 - 3x2 - 10

a2 + 7a + 6 a2 - 3a - 10

(a + 1)(a + 6) (a - 5)(a + 2)

(x2 + 1)(x2 + 6) (x2 - 5)(x2 + 2)

6­4 Solving Polynomial Equations Day 2

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January 14, 2009

Jan 6­7:00 AM

Solving a Higher-Degree Polynomial Equation:

Example #4: x4 - x2 = 12

x4 - x2 - 12 = 0

(x2 - 4)(x2 + 3) = 0

(x - 2)(x + 2)(x2 + 3) = 0

x = 2 or x = -2 or x2 = -3

x = ±2 or √x2 = ±√-3

x = ±2 or x = ±i√3

The solutions are 2, -2, i√3, and -i√3.

6­4 Solving Polynomial Equations Day 2

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January 14, 2009

Jan 6­7:01 AM

Example #5: x4 - 11x2 + 18 = 0

x4 - 11x2 + 18 = 0

(x2 - 2)(x2 - 9) = 0

x2 = 2 or (x - 3)(x + 3) = 0

√x2 = ±√2 or x - 3 = 0 or x + 3 = 0

x = ±√2 or x = ±3

The solutions are √2, -√2, 3, and -3.

6­4 Solving Polynomial Equations Day 2

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January 14, 2009

Jan 9­1:37 PM

Homework:

Practice 6-4