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Let \(t\) be a root of \(f(x) = x^3 - x + 2\). Evaluate \(t^6 - t^2 + 4t\).

 Mar 1, 2019
 #1
avatar+17331 
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The only root of x^3-x+2 is   -1.5214  (by graphing)

 

-1.5214^6 - ( -1.5214)^2 +4(-1.5214) = 4.00

 Mar 1, 2019
 #2
avatar+21848 
+1

Let t be a root of

\(f(x) = x^3 - x + 2\).

Evaluate

\(t^6 - t^2 + 4t\).

 

1.

\(\begin{array}{|rcll|} \hline f(t) = t^3 - t + 2 &=& 0 \\ t^3 - t + 2 &=& 0 \\ \mathbf{t^3} &\mathbf{=}&\mathbf{t-2} \\ \hline \end{array}\)

 

2.

\(\begin{array}{|rcll|} \hline t^6 - t^2 + 4t &=& (t^3)^2 - t^2 + 4t \quad | \quad \mathbf{t^3=t-2} \\ &=& (t-2)^2 - t^2 + 4t \\ &=& t^2-4t + 4 - t^2 + 4t \\ \mathbf{t^6 - t^2 + 4t} &\mathbf{=}&\mathbf{4} \\ \hline \end{array} \)

 

laugh

 Mar 1, 2019

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