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avatar+123 

a is a positive integer 

equation: x^3 + (a+17)x^2 + (38-a)x - 56 = 0 

The equation has 3 integer roots

what is the value of a?

what are the three roots?

 hint: x=1 is one of the roots

 Oct 25, 2018
 #1
avatar+98044 
+2

If 1 is a root...let's perform  some synthetic division

 

1  [  1     a + 17    38   - a      - 56  ]

                     1      a + 18        56

   _________________________

      1     a + 18        56            0

 

 

The remaining polynomial is    

 

x^2  + (a + 18)x  +  56

 

Factors of 56  are  1   2   4    7   8   14   28   56

 

Since a  is positive...only the pairs  2, 28    and  1, 56   will give us what we want

 

So...the possibiliites for the factorization of this are

 

(x +28) ( x + 2) =  x^2  + 30x  + 56       so  a  =  12       or

 

( x + 56) ( x + 1)  = x^2  + 57x  +  56    so   a  = 39

 

Checking this

 

x^3 + 29x^2 + 26x  - 56  = 0  produces the three roots of 1, -28  and  -2

 

And

 

x^3  + 56x^2 -1x - 56  = 0   produces the three roots   1, -1, - 56

 

 

 

cool cool cool

 Oct 25, 2018
 #2
avatar+123 
+1

thank you!!

hearts123  Oct 27, 2018

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