ch02 - Chapter 2 Polynomial Equations and Inequalities...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
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 .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 35

ch02 - Chapter 2 Polynomial Equations and Inequalities...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online