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Consider the polynomial $$ f(t) = t^3 + 6t^2 + 12t + 8. $$ What are the coefficients of $f(t-1)$? Enter your answer as an ordered list of four numbers. For example, if your answer were $f(t-1) = t^3 + 3t^2 - 2t + 7$, you'd enter (1,3,-2,7). (This is not the actual answer.)

 Apr 26, 2015

Best Answer 

 #2
avatar+126679 
+9

f(t-1)  = (t-1)^3 + 6(t-1)^2 + 12(t-1) + 8  =

[t^3 - 3t^2 + 3t - 1] + [6t^2 - 12t + 6] + [12t - 12 ] + 8 =

t^3 + 3t^2  + 3t + 1     =  (t + 1)^3

So...the coefficients  are  (1,3,3,1 )

 

  

 Apr 26, 2015
 #1
avatar+906 
0

i already feel so dumb right now

 Apr 26, 2015
 #2
avatar+126679 
+9
Best Answer

f(t-1)  = (t-1)^3 + 6(t-1)^2 + 12(t-1) + 8  =

[t^3 - 3t^2 + 3t - 1] + [6t^2 - 12t + 6] + [12t - 12 ] + 8 =

t^3 + 3t^2  + 3t + 1     =  (t + 1)^3

So...the coefficients  are  (1,3,3,1 )

 

  

CPhill Apr 26, 2015

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