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Synthetic division reveals another factor of 16x2 − 4x + 1. 8x3 − 1 = (2x − 1)(4x2 + 2x + 1) 64x3 + 1 = (4x + 1)(16x2 − 4x + 1) b) The expressions have the same signs. x + y is a factor of x3 + y 3 . The other factor is x2 − xy + y 2 . c) The expressions have the same signs. x − y is a factor of x3 − y 3 . The other factor is x2 + xy + y 2 . P (x) = 8x3 + 125 = (2x)3 + (5)3 d) = (2x + 5)((2x)2 − (2x)(5) + (5)2 ) = (2x + 5)(4x2 − 10x + 25) Q(x) = 27x3 − 64 = (3x)3 − (4)3 = (3x − 4)((3x)2 + (3x)(4) + (4)2 ) = (3x − 4)(9x2 + 12x + 16) P (x) = x6 + y 6 = (x2 )3 + (y 2 )3 e) = (x2 + y 2 )(x4 − x2 y 2 + y 4 ) f) Answers may vary.

2 h A = 6x2 − 5x − 4 and h = 3x − 4 Thus, 2(6x2 − 5x − 4) 3x − 4 2(3x − 4)(2x + 1) = 3x − 4 = 2(2x + 1) = 4x + 2 b= The base of the triangle is represented by 4x + 2. 2 Page 52 Question 15 The area of the trapezoid is given by A = 1 2A h(a + b), where h is the height and a and b are the bases. Thus, h = . 2 a+b A = 12y 2 − 11y + 2 a = 5y − 3 b = 3y + 1 Thus, 2(12y 2 − 11y + 2) (5y − 3) + (3y + 1) 2(3y − 2)(4y − 1) = 8y − 2 2(3y − 2)(4y − 1) = 2(4y − 1) = 3y − 2 h= The height of the trapezoid is represented by 3y − 2.

Y − 2 is a factor of P (y). Synthetic division reveals another factor of R(y) = 4y 2 + 4y + 1. Factor R(y) to obtain the remaining roots. = 32 − 16 − 14 − 2 =0 R(y) = 0 R(x) = 0 4y + 4y + 1 = 0 2 2x + x − 1 = 0 (2x − 1)(x + 1) = 0 1 x = or −1 2 2 (2y + 1)2 = 0 y=− 1 The roots are −1 and . 5 Page 81 Question 21 Let V (x) be the volume of the solid, in cubic centimetres. 1 2 1 The roots are 1, 2, and − . 2 V (x) = (x − 1)(x − 2)(x + 3) = (x − 1)(x2 + x − 6) = x3 − 7x + 6 Given a volume of 42 cm3 , we can solve for x.

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Calculus & Advanced Functions by Ryerson


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