# how to use barcode in c#.net Worked-Out Solutions to Exercises: s 21 to 29 in Software Painting QR Code in Software Worked-Out Solutions to Exercises: s 21 to 29

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and break it down to (x + 2)5( 3x + 1)5 = 0 (c) Here is the original equation: (4x 2 9)3 = 0 Here, the coefficient of x 2 is 4, the coefficient of x is 0, and the constant is 9 This trinomial factors into the product of (2x + 3) and (2x 3) Therefore, we can rewrite the original equation as [(2x + 3)(2x 3)]3 = 0 and break it down to (2x + 3)3(2x 3)3 = 0 4 In each case, we can remove the exponents from the binomials, and then consider each binomial separately as a first-degree equation when it is set equal to 0 The real roots of the higher-degree equation are equal to the solutions of the first-degree equations The multiplicity of each root is the power to which its associated binomial is raised (a) The real roots are found by solving x 1=0 and x 2=0 Those roots are x = 1 or x = 2 Because each binomial is squared, each of these roots has multiplicity 2 (b) The real roots are found by solving x+2=0 and 3x + 1 = 0 Those roots are x = 2 or x = 1/3 Because each binomial is raised to the fifth power, each of these roots has multiplicity 5 (c) The real roots are found by solving 2x + 3 = 0
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696 Worked-Out Solutions to Exercises: s 21 to 29
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and 2x 3 = 0 Those roots are x = 3/2 or x = 3/2 Because each binomial is cubed, each of these roots has multiplicity 3 5 Here s the original binomial factor equation, which we have been told to solve and scrutinize: (x 3/2)2(2x 7)2(7x)3( 3x + 5)5 = 0 Let s set each binomial equal to 0, and then solve the resulting first-degree equations Those equations are x 3/2 = 0 2x 7 = 0 7x = 0 3x + 5 = 0 The solutions to these first-degree equations, and therefore the real roots of the higherdegree equation, are x = 3/2 or x = 7/2 or x=0 or x = 5/3
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The solution set is X = {3/2, 7/2, 0, 5/3} The multiplicity of each root is the same as the power to which its binomial is raised in the original equation Therefore, the root x = 3/2 has multiplicity 2, the root x = 7/2 has multiplicity 2, the root x = 0 has multiplicity 3, and the root x = 5/3 has multiplicity 5 The degree of the original equation is the sum of the exponents attached to the factors, which is 2 + 2 +3 + 5 = 12 6 Here s the original binomial factor equation once again, for reference: (x + 4)(2x 8)2(x/3 + 12)3 = 0 Let s set each binomial equal to 0, and then solve the resulting first-degree equations Those equations are x+4=0 2x 8 = 0 x/3 + 12 = 0 The solution to the first of these is x = 4 The solution to the second is x = 4 To solve the third equation, we can subtract 12 from each side and then multiply through by 3, obtaining x = 36 The roots of the original equation are therefore x = 4 or x=4 or x = 36
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The solution set is X = { 4, 4, 36} The multiplicity of each root is the same as the power to which its binomial is raised in the original equation Therefore, the root x = 4 has multiplicity 1, the root x = 4 has multiplicity 2, and the root x = 36 has multiplicity 3 The degree of the original equation is the sum of the exponents attached to the factors, which is 1 + 2 + 3 = 6 7 For reference, the polynomial equation is 2x 5 3x 3 2x + 2 = 0 We have many options! The largest absolute value of any coefficient or constant is 3, so we can try 3 for the upper bound and 3 for the lower bound If either or both of these fail, we can try values farther from 0 The coefficients and constant, in order of decreasing powers of x, are 2, 0, 3, 0, 2, and 2 (The coefficients of x 4 and x 2 are both equal to 0) Here s the synthetic division array for the test root 3:
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