+0  
 
0
38
4
avatar+1159 

Help.

NotSoSmart  Dec 1, 2017
Sort: 

4+0 Answers

 #1
avatar+79819 
+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

CPhill  Dec 1, 2017
 #2
avatar+79819 
+1

2. 

 

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

 

7  - √3    and     2  + √ 6

 

 

cool cool cool

CPhill  Dec 1, 2017
 #3
avatar+79819 
+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

CPhill  Dec 1, 2017
 #4
avatar+79819 
+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

CPhill  Dec 1, 2017

4 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details