Explanation Given Information: The provided polynomial equation is x 3 + 6 x 2 + 11 x + 6 = 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 + 6 x 2 + 11 x + 6 = 0 . First determine the single solution in this cubic equation x 3 + 6 x 2 + 11 x + 6 = 0 Compare the equation x 3 + 6 x 2 + 11 x + 6 = 0 with the standard cubic equation a x 3 + b x 2 + c x + d = 0 a = 1 and d = 6 Therefore, possible factor of 1 is x = ± 1 and 6 is x = ± 1 , x = ± 2 , x = ± 3 and x = ± 6 . First try x = − 1 substitute in the provided polynomial x 3 + 6 x 2 + 11 x + 6 = 0 , ( − 1 ) 3 + 6 ( − 1 ) 2 + 11 ( − 1 ) + 6 = ? 0 − 1 + 6 − 11 + 6 = ? 0 0 = 0 Which is true. Therefore, x + 1 is a factor of the polynomial x 3 + 6 x 2 + 11 x + 6 = 0 . In order to find other two values of x , divide x 3 + 6 x 2 + 11 x + 6 = 0 by x + 1 , x + 1 x 2 + 5 x + 6 x 3 + 6 x 2 + 11 x + 6 x 3 + x 2 _ 5 x 2 + 11 x 5 x 2 + 5 x _ 6 x + 6 6 x + 6 _ 0 Thus, x 3

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ISBN: 9781337280426

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 0.5, Problem 31E

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To calculate: The all real solutions in the polynomial equation x3+6x2+11x+6=0.

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Chapter 0.5, Problem 30E

Chapter 0.5, Problem 32E

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The all real solutions in the polynomial equation x 3 + 6 x 2 + 11 x + 6 = 0 .

Solution Summary: The author calculates the real solutions in the polynomial equation x3+6, which has three distinct solutions.

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To calculate: The all real solutions in the polynomial equation x3+6x2+11x+6=0.

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