Explanation Given Information: The provided polynomial equation is x 3 − x 2 − 5 x + 5 = 0 . Formula used: Zero product rule: An equation a b = 0 is true if and only if a = 0 or b = 0 , both A product is zero if and only if at least one factor is zero The value of x in the equation a x 2 + b x + c = 0 can be given by the quadratic formula, x = − b ± b 2 − 4 a c 2 a where a , b and c are real numbers with a ≠ 0 . Calculation: Consider the polynomial equation x 3 − x 2 − 5 x + 5 = 0 . First determine the single solution in this cubic equation x 3 − x 2 − 5 x + 5 = 0 Compare the equation x 3 − x 2 − 5 x + 5 = 0 with the standard cubic equation a x 3 + b x 2 + c x + d = 0 . a = 1 and d = 5 Therefore, possible factor of 1 is x = ± 1 and 5 is x = ± 1 , x = ± 5 . First try x = 1 substitute in the provided polynomial equation x 3 − x 2 − 5 x + 5 = 0 , ( 1 ) 3 − ( 1 ) 2 − 5 ( 1 ) + 5 = ? 0 1 − 1 − 5 + 5 = ? 0 0 = 0 Which is true. Therefore, x − 1 is a factor of the polynomial x 3 − x 2 − 5 x + 5 = 0 . In order to find other two values of x , divide x 3 − x 2 − 5 x + 5 = 0 by x − 1 , x − 1 x 2 − 5 x 3 − x 2 − 5 x + 5 x 3 − x 2 _ − 5 x + 5 − 5 x + 5 _ 0 Thus, x 3 − x 2 − 5 x + 5 = ( x − 1 ) ( x 2 − 5 ) Therefore, ( x − 1 ) ( x 2 − 5 ) = 0 Apply the principle of zero-product, Either x − 1 = 0 or x 2 − 5 = 0 If x − 1 = 0 then, Add 1 on both sides, x − 1 + 1 = 0 + 1 x + 0 = 1 x = 1 If ( x 2 − 5 ) = 0 then Compare the above equation with the equation a x 2 + b x + c = 0 , a = 1 , b = 0 and c = − 5 Put the values of a , b , c in the quadratic formula x = − b ± b 2 − 4 a c 2 a

7th Edition

ISBN: 9781337291248

Author: Waner, Stefan.

Publisher: Cengage Learning,

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Chapter 0.5, Problem 39E

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Chapter 0.5, Problem 38E

Chapter 0.5, Problem 40E

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Exercises 105-120: Complete the following. (a) Write the equation as ax² + bx + e = 0 with a > 0. (b) Calculate the discriminant b² – 4ac and determine the number of real solutions. (c) Solve the equation. 105. 3x² = 12 106. 8x - 2 = 14 107. x² – 2x = -1 108. 6x² = 4x 109. 4x = x? 110. 16x + 9 = 24x 111. x² + 1 = x 112. 2x² + x = 2 113. 2x² + 3x = 12 – 2x 114. 3x² + 3 = 5x 115. x(x – 4) = -4 116. + 3x = x – 4 117. x(x + 2) = -13 118. 4x = 6 + x? 119. 3x = 1- x 120. x(5x – 3) = 1

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Applied Calculus

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Ch. 0.1 - Prob. 11E Ch. 0.1 - Prob. 12E Ch. 0.1 - Prob. 13E Ch. 0.1 - Calculate each expression in Exercises 124, giving...Ch. 0.1 - Prob. 15E Ch. 0.1 - Prob. 16E Ch. 0.1 - Calculate each expression in Exercises 124, giving...Ch. 0.1 - Prob. 18E Ch. 0.1 - Prob. 19E Ch. 0.1 - Prob. 20E Ch. 0.1 - Prob. 21E Ch. 0.1 - Prob. 22E Ch. 0.1 - Prob. 23E Ch. 0.1 - Prob. 24E Ch. 0.1 - Prob. 25E Ch. 0.1 - Prob. 26E Ch. 0.1 - Prob. 27E Ch. 0.1 - Prob. 28E Ch. 0.1 - Prob. 29E Ch. 0.1 - Prob. 30E Ch. 0.1 - Prob. 31E Ch. 0.1 - Prob. 32E Ch. 0.1 - Prob. 33E Ch. 0.1 - Prob. 34E Ch. 0.1 - Prob. 35E Ch. 0.1 - Prob. 36E Ch. 0.1 - Convert each expression in Exercises 25-50 into...Ch. 0.1 - Prob. 38E Ch. 0.1 - Prob. 39E Ch. 0.1 - Prob. 40E Ch. 0.1 - Prob. 41E Ch. 0.1 - Prob. 42E Ch. 0.1 - Prob. 43E Ch. 0.1 - Prob. 44E Ch. 0.1 - Prob. 45E Ch. 0.1 - Prob. 46E Ch. 0.1 - Prob. 47E Ch. 0.1 - Prob. 48E Ch. 0.1 - Prob. 49E Ch. 0.1 - Prob. 50E Ch. 0.2 - Evaluate the expression sin Exercises 116. 33 Ch. 0.2 - Prob. 2E Ch. 0.2 - Evaluate the expression sin Exercises 116. 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In Exercises 2946,...Ch. 0.8 - Prob. 32E Ch. 0.8 - Let a=log2,b=log3, and c=log7. In Exercises 2946,...Ch. 0.8 - Prob. 34E Ch. 0.8 - Prob. 35E Ch. 0.8 - Prob. 36E Ch. 0.8 - Prob. 37E Ch. 0.8 - Prob. 38E Ch. 0.8 - Prob. 39E Ch. 0.8 - Prob. 40E Ch. 0.8 - Prob. 41E Ch. 0.8 - Prob. 42E Ch. 0.8 - Prob. 43E Ch. 0.8 - Prob. 44E Ch. 0.8 - Prob. 45E Ch. 0.8 - Prob. 46E Ch. 0.8 - Prob. 47E Ch. 0.8 - Prob. 48E Ch. 0.8 - Prob. 49E Ch. 0.8 - Prob. 50E Ch. 0.8 - Prob. 51E Ch. 0.8 - Prob. 52E Ch. 0.8 - Prob. 53E Ch. 0.8 - Prob. 54E Ch. 0.8 - Prob. 55E Ch. 0.8 - Prob. 56E

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The all real solutions in the polynomial equation x 3 − x 2 − 5 x + 5 = 0 .

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Chapter 0 Solutions