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Chart Paper Problem #6:

 Feb 19, 2019
edited by GAMEMASTERX40  Feb 19, 2019
 #1
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+4

x^3 + 2x^2 - 7x - 2

 

We have something called the Rational Roots Theorem here that might help us....without going into detail ...the possible  rational roots are   1, - 1 , 2 or   -2

 

I see that 1 is not a root

I see that -1  isn't either

But 2 is a root because  (2)^3 + 2(2)^2  -7(2) - 2  =  8 + 8 - 14 - 2   = 0

 

So....using synthetic division, we can find  the remaining polynomial

 

 

2   [    1    2    - 7    -2 ]

                2     8     2

______________________

         1     4     1      0

 

 

The remaining polynomial is   x^2 + 4x + 1

 

We can complete the square on x to find the other two roots

 

x^2 + 4x + 4 =  - 1 + 4

 

(x + 2)^2 =  3    take both roots

 

x + 2 = ±√ 3      subtract 2 from both sides and we get the other two solutions

 

x = √3  - 2     or   x  = -√3   -  2

 

 

 

cool cool cool

 Feb 19, 2019

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