B graphing p x using the window suggested in part a

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(B) Graphing P ( x ) using the window suggested in part (A), we obtain –30000 –10 100000 20 The real zeros are -7.47 and 14.03.
228 CHAPTER 4 POLYNOMIAL AND RATIONAL FUNCTIONS 63. (A) We form a synthetic division table. 4 -40 -1,475 7,875 -10,000 0 10 20 30 -10 -20 4 -40 -1,475 7,875 -10,000 4 0 -1,475 -6,875 -78,750 4 40 -675 -5,625 -122,500 4 80 925 35,625 1,058,750 4 -80 -675 14,625 -156,250 4 -120 925 -10,625 202,500 From the table, 30 is an upper bound and -20 is a lower bound. (B) Graphing P ( x ) using the window suggested in part (A), we obtain the graph shown at the right. It appears that there are zeros between -20 and -10, and between 20 and 30, and there may be a zero of even multiplicity between 0 and 10. –200000 –20 500000 30 Applying zero and maximum routines, we obtain -17.66 and 20.66 are simple zeros and 2.50 is probably a double zero. To check that it is really a zero, we note 4 -40 -1475 7875 -10000 10 -75 -3875 10000 2.50 4 -30 -1550 4000 0 Thus 2.50 is definitely a zero. You can check that it is a double zero using a second synthetic division. 65. (A) We form a synthetic division table. 0.01 -0.1 -12 0 0 9,000 0 10 20 30 40 -10 -20 -30 -40 0.01 -0.1 -12 0 0 9,000 0.01 0 -12 -120 -1,200 -3,000 0.01 0.1 -10 -200 -4,000 -71,000 0.01 0.2 -6 -180 -5,400 -153,000 0.01 0.3 0 0 0 9,000 0.01 -0.2 -10 100 -1,000 19,000 0.01 -0.3 -6 120 -2,400 57,000 0.01 -0.4 0 0 0 9,000 0.01 -0.5 8 -320 12,800 -503,000 From the table, 40 is an upper bound and -40 is a lower bound.
SECTION 4-2 229 (B) Graphing P ( x ) using a window suggested by part (A), we obtain –300000 –40 100000 50 The real zeros are -30.45, 9.06, and 39.80. A graph in a smaller window confirms that there are no other real zeros between -10 and 10, and therefore most likely no other real zeros. –5000 –10 20000 10 67. Yes. According to the Remainder Theorem, if the remainder is 10 when P ( x ) is divided by x + 4, then P (-4) = 10. Similarly, if the remainder is -8 when dividing by x + 5, then P (-5) = -8. Since P (-5) and P (-4) have opposite signs, there must be a zero between -5 and -4. 69. If all coefficients of P ( x ) are nonnegative, then no positive replacement of x could possibly make P ( x ) = 0; hence, there are no positive zeros of P ( x ). All zeros must therefore be less than or equal to 0, and 0 is an upper bound for the zeros. 71. Let ( x, x 2 ) be a point on the graph of y = x 2 . Then the distance from (1, 2) to ( x, x 2 ) must equal 1 unit. Applying the distance formula, we have, ( x ± 1) 2 + ( x 2 ± 2) 2 = 1 ( x - 1) 2 + ( x 2 - 2) 2 = 1 x 2 - 2 x + 1 + x 4 - 4 x 2 + 4 = 1 x 4 - 3 x 2 - 2 x + 4 = 0 Let P ( x ) = x 4 - 3 x 2 - 2 x + 4 The only rational zero is 1. 1 0 -3 -2 4 1 1 -2 -4 1 1 1 -2 -4 0 P ( x ) = ( x - 1)( x 3 + x 2 - 2 x - 4) We examine Q ( x ) = x 3 + x 2 - 2 x - 4. Graphing y = Q ( x ) we obtain the graph shown at the right. Q ( x ) (and P ( x )) has a second zero at x = 1.7. Therefore, the two real zeros of P ( x ) are 1 and 1.7. Hence the two required points are (1, 1) and (1.7, 1.7 2 ) = (1.7, 2.9). –10 –10 10 10
230 CHAPTER 4 POLYNOMIAL AND RATIONAL FUNCTIONS 73. x x 18 in 24 in 24 - 2 x x 18 - 2 x 24 - 2x 18 - 2x From the above figures it should be clear that V = length ± width ± height = (24 - 2 x )(18 - 2 x ) x 0 < x < 9 (why?) We solve (24 - 2 x )(18 - 2 x ) x = 600 432 x - 84 x 2 + 4 x 3 = 600 4 x 3 - 84 x 2 + 432 x - 600 = 0 x 3 - 21 x 2 + 108 x - 150 = 0 Let P ( x ) = x 3 - 21 x 2 + 108 x - 150. Graphing y = P ( x ) on the interval 0 < x < 9, we obtain –30 0 30 9 The zeros of P ( x ) are 2.3 and 4.6, 0 < x < 9.