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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 150 – 152 Thus, (3 x 3 – 4 x 2 + 11 x – 2) ( 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 2 5 = 0, 5 x – 2 is a factor. By long division, 5 x 3 + 8 x 2 + 21 x – 10 = (5 x – 2)( x 2 + 2 x + 5). The expression x 2 + 2 x + 5 does not factor in x R .
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