Explanation Given Information: The provided polynomial equation is y 3 − 2 y 2 − 2 y − 3 = 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 y 3 − 2 y 2 − 2 y − 3 = 0 . First determine the single solution in this cubic equation y 3 − 2 y 2 − 2 y − 3 = 0 Compare the equation y 3 − 2 y 2 − 2 y − 3 = 0 with the standard cubic equation a x 3 + b x 2 + c x + d = 0 . a = 1 and d = − 3 Therefore, possible factor of 1 is x = ± 1 and 3 is x = ± 1 , x = ± 2 and x = ± 3 . First try y = 3 substitute in the provided polynomial equation y 3 − 2 y 2 − 2 y − 3 = 0 , ( 3 ) 3 − 2 ( 3 ) 2 − 2 ( 3 ) − 3 = ? 0 27 − 18 − 6 − 3 = ? 0 0 = 0 Which is true Therefore, y − 3 is a factor of the polynomial y 3 − 2 y 2 − 2 y − 3 = 0 . In order to find other two values of x , divide y 3 − 2 y 2 − 2 y − 3 = 0 by y − 3 , y − 3 y 2 + y + 1 y 3 − 2 y 2 − 2 y − 3 y 3 − 3 y 2 _ y 2 − 2 y &#

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Author: Waner

Publisher: CENGAGE L

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Chapter 0.5, Problem 38E

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To calculate: The all real solutions in the polynomial equation y3−2y2−2y−3=0.

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Chapter 0.5, Problem 37E

Chapter 0.5, Problem 39E

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The all real solutions in the polynomial equation y 3 − 2 y 2 − 2 y − 3 = 0 .

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