Explanation Given Information: The provided polynomial equation is ( x 2 + 3 x + 2 ) ( x 2 − 5 x + 6 ) = 0 . Formula used: Principle of 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 − 4 a c 2 a where a , b and c are real numbers. Calculation: Consider the polynomial equation ( x 2 + 3 x + 2 ) ( x 2 − 5 x + 6 ) = 0 . Apply the zero-product rule, If x 2 + 3 x + 2 = 0 Compare the above equation with the equation a x 2 + b x + c = 0 , a = 1 , b = 3 and c = 2 Put the values of a , b , c in the quadratic formula x = − b ± b 2 − 4 a c 2 a . x = − 3 ± ( 3 ) 2 − 4 ⋅ 1 ⋅ 2 2 ⋅ 1 = − 3 ± 9 − 8 2 = − 3 ± 1 2 = − 3 ± 1 2 Further simplify, x =

7th Edition

ISBN: 9781337291248

Author: Waner, Stefan.

Publisher: Cengage Learning,

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Chapter 0.5, Problem 43E

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To calculate: The all real solutions in the polynomial equation (x2+3x+2)(x2−5x+6)=0.

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Chapter 0.5, Problem 42E

Chapter 0.5, Problem 44E

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Exercises 38–40 will help you prepare for the material covered in the first section of the next chapter. In Exercises 38-39, simplify each algebraic expression. 38. (-9x³ + 7x? - 5x + 3) + (13x + 2r? – &x – 6) 39. (7x3 – 8x? + 9x – 6) – (2x – 6x? – 3x + 9) 40. The figures show the graphs of two functions. y y 201 10- .... -20- flx) = x³ glx) = -0.3x + 4x + 2

For Exercises 115–120, factor the expressions over the set of complex numbers. For assistance, consider these examples. • In Section R.3 we saw that some expressions factor over the set of integers. For example: x - 4 = (x + 2)(x – 2). • Some expressions factor over the set of irrational numbers. For example: - 5 = (x + V5)(x – V5). To factor an expression such as x + 4, we need to factor over the set of complex numbers. For example, verify that x + 4 = (x + 2i)(x – 2i). 115. а. х - 9 116. а. х? - 100 117. а. х - 64 b. x + 9 b. + 100 b. x + 64 118. а. х — 25 119. а. х— 3 120. а. х — 11 b. x + 25 b. x + 3 b. x + 11

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Chapter 0 Solutions

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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The all real solutions in the polynomial equation ( x 2 + 3 x + 2 ) ( x 2 − 5 x + 6 ) = 0 .

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To calculate: The all real solutions in the polynomial equation (x2+3x+2)(x2−5x+6)=0.

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Chapter 0 Solutions