Calculus & Advanced Functions by Ryerson

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Let V (x) be the volume of the box, in cubic centimetres. V (x) = (x + 5)(x + 10)(x + 10) = (x + 5)(x2 + 20x + 100) = x3 + 25x2 + 200x + 500 Given a volume of 1008 cm3 , the value of x can be determined. V (x) = 1008 x + 25x + 200x + 500 = 1008 x3 + 25x2 + 200x − 508 = 0 Let P (x) = x3 + 25x2 + 200x − 508. P (2) = 23 + 25(2)2 + 200(2) − 508 = 8 + 100 + 400 − 508 =0 3 2 Thus, x = 2 is a root of P (x) = 0. From x = 2, the dimensions of the new prism are 7 cm by 12 cm by 12 cm.

5 Page 80 Question 13 a) The base radius of the cone can be expressed as r = h = x + 10. 100 − x2 . The height of the cone can be expressed as 1 2 πr h 3 2 1 100 − x2 , x + 10 = π 100 − x2 (x + 10) 3 1 V (x) = π(100 − x2 )(x + 10) 3 V (r, h) = V b) V (x) = 1000 GRAPHING CALCULATOR 1 π(100 − x2 )(x + 10) = 1000 3 1 π(100x + 1000 − x3 − 10x2 ) − 1000 = 0 3 . 49 . 49 m. 61 m. 49 m. c) The dimensions of the pyramid are x by x by x + 2. Let V be the volume. V = 1000 GRAPHING CALCULATOR 1 2 x (x + 2) = 1000 3 1 3 (x + 2x2 ) − 1000 = 0 3 .

D) Factor by grouping. P (x) = 8x3 + 4x2 − 2x − 1 P (x) = 8x3 − 12x2 − 2x + 3 = 4x2 (2x + 1) − 1(2x + 1) = 4x2 (2x − 3) − 1(2x − 3) = (2x + 1)(4x2 − 1) = (2x − 3)(4x2 − 1) = (2x + 1)(2x + 1)(2x − 1) = (2x − 3)(2x + 1)(2x − 1) = (2x + 1) (2x − 1) 2 So, 8x3 − 12x2 − 2x + 3 = (2x − 3)(2x + 1)(2x − 1). So, 8x3 + 4x2 − 2x − 1 = (2x + 1)2 (2x − 1). 4 Page 69 Question 13 P (x) = x4 + 6x3 + 11x2 + 6x = x(x3 + 6x2 + 11x + 6) = xQ(x) Q(−1) = (−1)3 + 6(−1)2 + 11(−1) + 6 = −1 + 6 − 11 + 6 =0 To this point, x and x + 1 are factors of P (x).

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