Question 1028785
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#1 Factor the expression on the left side of the equation. Then solve the equation.
 x^3 - 2x^2 - 5x = 0
 		
A. 0, -1 +/- sqrt 24
B. -5, 0, 3
C. 0, 1 +/- sqrt 6
D. 0, -1 +/- sqrt 6
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Factor:

{{{x*(x^2 - 2x -5)}}} = {{{0}}}.

Solve the equation {{{(x^2 - 2x -5)}}} = {{{0}}}. Apply the quadratic formula

{{{x[1,2]}}} = {{{(2 +- sqrt(4 + 4*5))/2}}} = {{{(2 +- sqrt(24))/2}}} = {{{(2 +- 2*sqrt(6))/2}}} = {{{1 +- sqrt(6)}}}.

The roots of the original equation are  0,  {{{1 +sqrt(6)}}}  and  {{{1 - sqrt(6)}}}.

The answer is option C.
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#2 Factor the expression on the left side of the equation. Then solve the equation.
x6 + 128x3 + 4096 = 0

A. 4, 2 +/- 2i*sqrt 3
B. 4 multiplicity of 2, 2 +/- 2i*sqrt3 multiplicity of 2
C. -4, 2 +/- 2i*sqrt 3
D. - 4 multiplicity of 2, 2 +/- 2i*sqrt3 multiplicity of 2


<pre>
Factor

{{{x6 + 128x3 + 4096}}} = {{{(x^3 + 64)^2}}}.   (1)

Now solve  

{{{(x^3 + 64)^2}}} = {{{0}}}.        (2)

It is reduced to 

{{{x^3 + 64)}}} = {{{0}}},         (3)

{{{x^3}}} = {{{-64}}},

{{{x[1,2,3]}}} = {{{-4}}},  {{{(-4)*(cos(2pi/3) + i*sin(2pi/3))}}},  {{{(-4)*(cos(2pi/3) - i*sin(2pi/3))}}},   or, which is the same 

{{{-4}}},  {{{(-4)*((-1/2) + i*(sqrt(3)/2))}}},  {{{(-4)*((-1/2) - i*(sqrt(3)/2))}}},   or,  which is the same again

{{{-4}}},  {{{2 +- 2i*sqrt(3)}}}.     (4).


Thus the solution of (3) are the roots (4).

Hence, the solution of (1) are the roots listed under the option D)
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