# ch02 - Chapter 2 Polynomial Equations and Inequalities...

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Review of Prerequisite Skills 1. b. There is no solution. 2. c. d. 3. b. d. 4 . c. d. f 1 2 1 2 2 1 2 4 1 2 5 53 8 32 = + + = f –– 33 2 3 4 3 5 52 () = () () +() + = fx x x x () =+ + 24 5 f 1 2 2 1 2 3 1 2 1 0 2 = + = f 22 2 3 21 15 2 = () () + = fx x x 23 1 2 -3 -2 -1 0 123 47 91 7 51 0 2 xx x x +< + < > -6 -5 -4 -3 -2 -1 0 45 2 7 4 29 9 2 x x x –. or 7 37 36 7 1 31 1 02 2 x += = 5. d. f. h. 6. e. f. h. 7. b. y = ± () ( ) = ± = 55 4 3 4 57 3 6 06 2 –– – ˙ .– . or 35 4 0 2 yy = x xxx == + = === 03 0 3 or or or or x 3 2 90 0 = = + = 74 0 1 0 4 7 1 = or or 73 4 0 10 2 + = 50 30 53 = or or 2 0 = + = 2 34 322 3 2 x = = + x x x 2 56 56 87 + 5 5 5 3 2 x = = + Chapter 2: Polynomial Equations and Inequalities 17 Chapter 2 • Polynomial Equations and Inequalities

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c. e. Solution 1 Solution 2 g. i. x = ± () ( ) ( ) () = ± = 99 4 1 6 21 95 7 2 83 07 2 –– ˙ .. or 25 2 3 0 6 96 0 22 2 xx x x x x =+ =− += p i = ±() ( ) ( ) = ± = ± 33 4 2 5 1 4 1 4 2 23 5 0 2 pp 21 30 210 30 3 1 2 3 + ( ) = = == or or or x = ( ) = ± = 55 4 2 3 54 9 4 5 2 –. or 3 0 2 = x = ( ) = ± = 4 1 4 1 2 57 2 –– – ˙ .– . or 0 2 ++= Exercise 2.1 2. b. The other factors can be found by dividing x – 5 into f ( x ) then checking factors for the quotient, either by inspection or using the Factor Theorem. 3. If then the factors are , and . This is true since f ( a ) is the remainder, and in this case, all remainders are zero, giving division that is complete. Also, this is the Factor Theorem. 4. a. x – 1 is a factor of only if d. Therefore, is not a factor of f ( x ). f. Therefore, is a factor of ( x ). x fx x x x f = + = 468 3 1 2 4 1 2 6 1 2 8 1 2 3 1 2 3 2 43 0 32 x –3 fx x x x f + =+() () + 62 3 336 3 3 0 Since then isa factor. 117 160 1 2 () = () –, f 10 () = . x 2 76 x + 3 + 12 ,– and ff f . 3 0 () = () =() = x x 6 18 Chapter 2: Polynomial Equations and Inequalities
5. a. b. x – 3 is a linear factor of f ( x ). c. The quadratic factor is 6. a. is the linear factor of f ( x ). c. By comparing coefficients, the quadratic factor is . 7. a. ( x – 1) is a factor of f ( x ). Comparing coefficients, ∴+ = () + xx x 32 43 1 3 –– . k = 1 . k –1 0 = k x xkx kx 1 3 13 3 += + =+ + + Let fx x x f () = () + = 3 3 114 0 –. 2 41 + k = 4 k 22 xxx x x k x xk 2 25 6 2 1 21 2 += + ()++ =++ + x + 2 g 2 2 5 2 6 0 = () () () + = gx x x x 6 2 1 ++ . ) x x x 2 3 1 3 2 3 3 3 0 2 2 2 f 332 3 2 33 27 18 6 3 0 3 2 () = () () = = ––– fx () = 3 ( x – 1) is a factor of f ( x ). = x 3 + ( k – 1) x 2 + … Comparing coefficients, k = 3 e. is a factor of . By dividing, . g. Because the function is quartic and the constant is –12, which presents many possibilities, we use the graphing calculator in mode in the function to establish .

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## This note was uploaded on 02/10/2010 for the course MATHS MCV-01 taught by Professor Mr.m during the Spring '10 term at Seneca.

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ch02 - Chapter 2 Polynomial Equations and Inequalities...

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