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To solve equations of degree 2, we can use factoring or use the quadratic formula.

For equations of higher degree, we can use the factoring method or the Factor Theorem, or graphing technology.

1.6 Solving Polynomial Equations

Roots of Polynomial Equations A Polynomial equation of degree n will have between 0 and n roots if n is even and between 1 and n roots if n is odd.

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1. Solve for x: x3 – 3x2 – 4x + 12 = 0(factor by grouping): x2(x – 3) – 4(x – 3) = 0

(x – 3)(x2 – 4) = 0

x = 3 or x = 2 or x = –2

(x – 3)(x – 2)(x + 2) = 0(diff. of squares)

Example:

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2. Solve for x: 2x3 – 7x2 + 7x – 2 = 0P(x) = 2x3 – 7x2 + 7x – 2

P(1) = 2(1)3 – 7(1)2 + 7(1) – 2

P(1) = 2 – 7 + 7 – 2

P(1) = 0 x – 1 is a factor

(x = 1 is one root of the equation)

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Divide: 2x3 – 7x2 + 7x – 2 by x – 12 – 7 + 7 – 2 1

2 2– 5

– 5 2

20

(2x2 – 5x + 2)

2x3 – 7x2 + 7x – 2 = 0

(x – 1)(2x – 1)(x – 2) = 0 11, , 22

x

Solution:

= (2x – 1)(x – 2)

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3. Solve for x: x3 + x – 10 = 0

P(x) = x3 + x – 10 P(2) = (2)3 + 2 – 10

P(2) = 8 + 2 – 10

P(2) = 0

x – 2 is a factor

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#3(cont’t) divide: x3 + x – 10 by x – 2

1 0 1 – 10 2

1 2 2

4 5

100

( x2 + 2x + 5)

x3 + x – 10 = (x – 2)(x2 + 2x + 5)

Use the quadratic formula

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a = 1 b = 2 c = 5

2 42

x cb b aa

2 (1)2 (2 5(1)

)42

x

1 2x i

#3 (cont’d): x2 + 2x + 5 = 0

2 4 202

x

2 162

x

2 162ix

The roots are x = 2 or

1 2x i

2 42

ix