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Help.

 Dec 1, 2017
 #1
avatar+98196 
+1

x^4  + 8x^3  + 7x^2 - 40x  - 60  = 0 

 

Write as

 

 x^4 + 8x^3 +  [12x^2   -  5x^2]  - 40x - 60 ]  = 0

 

x^2 [x^2 + 8x + 12 ]  -  [ 5x^2 + 40x + 60 ]  = 0

 

x^2 [  (x + 6) (x + 2) ]  -  [ (5x + 10) (x + 6) ]  = 0

 

x^2 [ (x + 6) (x + 2) ] - [ 5 ( x + 2)  (x + 6) ]  = 0

 

[ (x + 2) (x + 6) ] [ x^2  - 5 ]  = 0

 

(x + 2) (x + 6) (x^2 - 5)  = 0

 

The rational  roots will come from setting the first two linear factors to 0 and solving for x

 

Thus.....   x  = -2    and x  = -6     are the rational roots

 

 

cool cool cool

 Dec 1, 2017
 #2
avatar+98196 
+1

2. 

 

If   7 + √3   and 2 -  √6   are roots....so are their conjugates....

 

7  - √3    and     2  + √ 6

 

 

cool cool cool

 Dec 1, 2017
 #3
avatar+98196 
+1

3.

 

2x^4 - 5x^3 + 53x^2 -125x  + 75  =  0

 

Write as

 

2x^4  -  5x^3  +  3x^2  + 50x^2 - 125x + 75  = 0

 

x^2 (2x^2 - 5x + 3))  +  25 (2x^2 - 5x + 3)  =  0

 

x^2  [  (2x - 3) (x - 1) ]  + 25 [ (2x^2 - 3) (x - 1) ]  = 0

 

[ (2x- 3) ( x - 1)]   [ (x ^2 + 25 ]   = 0

 

The   roots are     3/2, 1  and  ±5i

 

 

cool cool cool

 Dec 1, 2017
 #4
avatar+98196 
+1

(8v + s)^5   =

 

(s + 8v)^5

 

All  terms are positive  and we will have 6 of them

 

The first is  s^5      and the last is   (8v)^5  = 32768v^5

 

And the second term  is    C(5,1) * s^4 * 8v   = 5* s^4 * 8v  =  40s^4v

 

So....the last answer is correct.....NSS!!!!!

 

cool cool cool

 Dec 1, 2017

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