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`x^3+3x^2+5x+15`

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mommat88 | eNotes Newbie

Posted May 22, 2013 at 9:39 PM via web

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`x^3+3x^2+5x+15`

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted May 23, 2013 at 12:36 AM (Answer #2)

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Factor `x^3+3x^2+5x+15` :

Factor two terms at a time:

`x^2(x+3)+5(x+3)`

Now (x+3) is a common factor of both remaining terms and can be factored out:

`(x+3)(x^2+5)`

`x^2+5` does not factor in the reals -- it can be fully factored in the complex numbers:

`(x+3)(x+sqrt(5)i)(x-sqrt(5)i)`

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crmhaske | College Teacher | (Level 3) Associate Educator

Posted May 22, 2013 at 10:06 PM (Answer #1)

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The equation can be factored as follows:      

       x^2+5
x+3|x^3+3x^2+5x+15
       x^3+3x^2
               0x^2+5x + 15
                         5x + 15
                                   0

Therefore:

x^3+3x^2+5x+15=(x+3)(x^2+5)

There exists one real root at x=-3

f'(x)=3x^2+6x+5=0

a=3; b=6; c=5

b^2-4ac = 36-4(3)(5)=-24<0

Therefore there are no maximum or minimum points.

f''(x)=6x+6=0 -> x=-1

f''(0)=6>0 -> concave up for x>-1

f''(-2)=-6<0 -> concave down for x<-1

Therefore there is an inflection point at x=-1 where the graph changes from concave downwards to concave upwards

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