chs1-4review

chs1-4review - Cumulative Review Solutions Chapters 1 4 2....

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Chapters 1–4 2. The given function is a polynomial function of degree three. The x- intercept s are –1 and 2. Since –1 is a double root, the graph is tangent to the x -axis at x = –1. The y -intercept is 2. Since the coefficient of the x 3 term is negative, the graph goes from the second quadrant to the fourth quadrant. 4. b. 3 x 3 +9 x 2 –13 x 2 +11 x –13 x 2 –39 x 50 x –2 50 x 1 150 – 152 Thus, (3 x 3 –4 x 2 +11 x –2) 4 ( x +3) =3 x 2 –13 x +50– } x 1 + 52 3 } 7. Let f ( x )= x 3 + kx 2 –4 x + 12. Since x –3 is a factor of f ( x ), f (3) = 0. Thus, 27 + 9 k –12+12=0 ,and k = –3. 8. Let f ( x )= x 4 –2 x 3 +5 x 2 –6 x – 8. To determine whether or not x –2 is a factor of f ( x ), we evaluate f (2). f (2)=16–16+20–12–8=0 Since f (2)=0 , x –2 is a factor of f ( x ) y x 2 0 2 – 1 9. We use the Factor Theorem to determine other factors of the given polynomial. We know that for x p to be a factor, p must be a divisor of 6. Let f ( x )= x 3 –2 x 2 –5 x +6 . Since f (1)=1–2–5+6=0, x –1 is a factor of f ( x ). Since f (3)=27–18–15+6=0 , x –3 is a factor of f ( x ). Thus, x 3 –2 x 2 –5 x +6=( x + 2), ( x – 1), and ( x –3) . 10. d. Let f ( x )=5 x 3 +8 x 2 +21 x – 10. Since f 1 } 2 5 } 2 =0 ,5 x –2 is a factor. By long division, 5 x 3 +8 x 2 +21
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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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chs1-4review - Cumulative Review Solutions Chapters 1 4 2....

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