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

Bundle: Finite Mathematics, Loose-leaf Version, 7th + WebAssign Printed Access Card for Waner/Costenoble's Finite Mathematics, 7th Edition, Single-Term

7th Edition

ISBN: 9781337604949

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 0.5, Problem 39E

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To calculate: The all real solutions in the polynomial equation x3−x2−5x+5=0.

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Chapter 0.5, Problem 38E

Chapter 0.5, Problem 40E

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Bundle: Finite Mathematics, Loose-leaf Version, 7th + WebAssign Printed Access Card for Waner/Costenoble's Finite Mathematics, 7th Edition, Single-Term

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

Solution Summary: The author explains how to calculate the real solutions in the polynomial equation x3-2-5x+5=0.

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To calculate: The all real solutions in the polynomial equation x3−x2−5x+5=0.

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