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1. Let $f(x)$ be the polynomial \[f(x)=x^7-3x^3+2.\]If $g(x) = f(x + 1)$, what is the sum of the coefficients of $g(x)$?

 

2. Expand $(2z^2 + 5z - 6)(3z^3 - 2z + 1)$.

Guest Mar 30, 2018
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2+0 Answers

 #1
avatar+2269 
+2

2. Distribute parenthesis,  apply minus-plus rules, and solve to get \(6z^5+15z^4-22z^3-8z^2+17z-6\)

tertre  Mar 30, 2018
 #2
avatar+85673 
+1

f(x)  = x^7 - 3x^3  + 2

g(x)  = f(x + 1)   =  (x + 1)^7 - 3(x+ 1)^3 + 2

The sum of the the coefficients  of g(x)  =   

Sum of the entries in Row 7 of Pascal^s Triangle  =  2^7 -

3 * Sum of the entries in Row 3  of Pascal's Triangle  = 3*2^3  +

2 =

 

2^7 - 3*2^3  +  2   =   106

 

 

 

(2z^2 + 5z - 6)(3z^3 - 2z + 1)

 

2z^2(3z^3 - 2z + 1)  + 5z(3z^3 - 2z + 1)  - 6(3z^3 - 2z + 1)

6z^5 - 4z^3 + 2z^2  + 15z^4 - 10z^2 + 5z  - 18z^3 + 12z - 6    combine like terms

6z^5 + 15z^4 - 22z^3 - 8z^2 + 17z - 6

 

 

cool cool cool

CPhill  Mar 30, 2018

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