Explanation Given information: The equation, ( x 2 − 1 ) 2 ( x + 2 ) 3 − ( x 2 − 1 ) 3 ( x + 2 ) 2 = 0 . Formula used: Zero product property, If a b = 0 then either a = 0 or b = 0 . Formula for difference of two squares, ( a + b ) ( a − b ) = a 2 − b 2 Where, a and b are real numbers. Quadratic formula, The solution of the quadratic equation a x 2 + b x + c = 0 is, x = − b ± b 2 − 4 a c 2 a Calculation: Consider the expression, ( x 2 − 1 ) 2 ( x + 2 ) 3 − ( x 2 − 1 ) 3 ( x + 2 ) 2 = 0 The left side expression is ( x 2 − 1 ) 2 ( x + 2 ) 3 − ( x 2 − 1 ) 3 ( x + 2 ) 2 . As both the term contains the term ( x 2 − 1 ) 2 ( x + 2 ) 2 . Thus, the common factor is ( x 2 − 1 ) 2 ( x + 2 ) 2 . Now, factor out the common factor and simplify as below, ( x 2 − 1 ) 2 ( x + 2 ) 3 − ( x 2 − 1 ) 3 ( x + 2 ) 2 = ( x 2 − 1 ) 2 ( x + 2 ) 2 ( x + 2 − ( x 2 − 1 ) ) = ( x 2 − 1 ) 2 ( x + 2 ) 2 ( x + 2 − x 2 + 1 ) = ( x 2 − 1 ) 2 ( x + 2 ) 2 ( x + 3 − x 2 ) = ( x − 1 ) ( x + 1 ) ( x + 2 ) 2 ( x + 3 − x 2 ) Therefore, the factors of the expression ( x 2 − 1 ) 2 ( x + 2 ) 3 − ( x 2 − 1 ) 3 ( x + 2 ) 2 are ( x − 1 ) ( x + 1 ) ( x + 2 ) 2 ( x + 3 − x 2 )

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ISBN: 9781337291408

Author: Stefan Waner, Steven Costenoble

Publisher: Brooks Cole

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Chapter 0.6, Problem 14E

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To calculate: The solution of the equation (x2−1)2(x+2)3−(x2−1)3(x+2)2=0.

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Chapter 0.6, Problem 13E

Chapter 0.6, Problem 15E

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Chapter 0 Solutions

Applied Calculus, Loose-leaf Version

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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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The solution of the equation ( x 2 − 1 ) 2 ( x + 2 ) 3 − ( x 2 − 1 ) 3 ( x + 2 ) 2 = 0 .

Solution Summary: The author calculates the solution of the quadratic equation ax2+bx+c=0.

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To calculate: The solution of the equation (x2−1)2(x+2)3−(x2−1)3(x+2)2=0.

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Chapter 0 Solutions