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Problem 17 – Solution – pg. 3/3

Posted in Solutions for Posted Problems by Lia on 06/23/2011

Problem 17 – Solution: 12, 3

b)  To solve the equation  P(x) = 0 means to find its roots.  The highest power of {{}{x}{}} forP(x) is {{}{5}{}}, therefore P(x) has {{}{5}{}} roots. The expresion can be written as :

[4]   P(x) = {x^5} - 2{x^4} - 6{x^3} + m{x^2} + nx + p = (x - x_1)(x - x_2)(x - x_3)(x - x_4)(x - x_5) where x_1; x_2; x_3; x_4 and x_5 are the five roots of P(x).

from [1] we have:

[5]  P(x) = {x^5} - 2{x^4} - 6{x^3} + m{x^2} + nx + p = (x - 3)(x + 1)(x - 1)(ax^2 +bx +c)

By  replacing a, b, c, m, n and {{}{p}{}} calculated at point a) into [5] we have:

[6]  P(x) = {x^5} - 2{x^4} - 6{x^3} + 8{x^2} + 5x - 6 = (x - 3)(x + 1)(x - 1)(x^2 + x - 2)

From [6] we can see that x_1 = 3, x_2 = -1 and x_3 = 1. The roots x_4 and x_5 will be equal with the roots of the quadratic equation x^2 + x - 2 = 0.

x_{4, 5} = {{- 1 {pm} {sqrt{1 + 8}}}/2} = {{- 1 {pm} {sqrt{9}}}/2} = {{- 1 {pm} 3}/2}

So: x_4 = {{- 1 + 3}/2} =  1 and x_5 = {{- 1 - 3}/2} =  - 2

And the roots of P(x) = 0 are: tabular{11}{11}{{{x_1} = 3}}; tabular{11}{11}{{{x_2} = - 1}}; tabular{11}{11}{{{x_3} = 1}}; tabular{11}{11}{{{x_4} =1}} and tabular{11}{11}{{{x_5} = - 2}}.

 

Problem 17 – Solution: 12, 3

 

 

 

 

 

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