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Determine the coefficients in front x and x^3 (x^2+9x+3)(x^3-4x+7/x) 

 Jul 20, 2016
 #1
avatar+128656 
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 (x^2+9x+3)(x^3-4x+7/x)  =  

 

x^5 + 9 x^4 - x^3 - 36 x^2 - 5 x + 21/x + 63

 

I assume you can take it from here

 

 

 

 

 

cool cool cool

 Jul 20, 2016
 #2
avatar+78 
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no i can't.... :(

 Jul 20, 2016
 #3
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Simplify the following:
(x^2+9 x+3) (x^3-4 x+7/x)

 

Put each term in x^3-4 x+7/x over the common denominator x: x^3-4 x+7/x  =  x^4/x-(4 x^2)/x+7/x:
x^4/x-(4 x^2)/x+7/x (x^2+9 x+3)

 

x^4/x-(4 x^2)/x+7/x = (x^4-4 x^2+7)/x:
Answer: |  (x^4 - 4x^2 + 7)/x (x^2 + 9x + 3)=x^5 + 9x^4 - x^3 - 36x^2 - 5x + 21/x + 63

 Jul 20, 2016

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