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If k is constant, what is the value of k so that the polynomial \(k^2x^3-8kx+16\) is divisible by x-1?

 Apr 30, 2018

Best Answer 

 #1
avatar+21819 
+2

If k is constant, what is the value of k so that the polynomial

\(k^2x^3-8kx+16\) is divisible by \(x-1\)?

 

\(k^2x^3-8kx+16 = k^2(x-1)(\ldots )(\ldots) \)

The first root is 1.

 

\(\begin{array}{|rcll|} \hline k^2 \cdot 1^3 - 8k \cdot 1 + 16 &=& 0 \\ k^2 - 8k + 16 &=& 0 \\ k &=& \dfrac{ 8\pm \sqrt{64-4\cdot 16} }{2} \\ k &=& \dfrac{ 8\pm 0 }{2} \\ k &=& \dfrac{ 8\pm 0 }{2} \\ \mathbf{k} & \mathbf{=} & \mathbf{4} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline 16x^3-32x+16 : (x-1) = 16x^2+16x-16 \\ \hline \end{array}\)

 

 

laugh

 Apr 30, 2018
 #1
avatar+21819 
+2
Best Answer

If k is constant, what is the value of k so that the polynomial

\(k^2x^3-8kx+16\) is divisible by \(x-1\)?

 

\(k^2x^3-8kx+16 = k^2(x-1)(\ldots )(\ldots) \)

The first root is 1.

 

\(\begin{array}{|rcll|} \hline k^2 \cdot 1^3 - 8k \cdot 1 + 16 &=& 0 \\ k^2 - 8k + 16 &=& 0 \\ k &=& \dfrac{ 8\pm \sqrt{64-4\cdot 16} }{2} \\ k &=& \dfrac{ 8\pm 0 }{2} \\ k &=& \dfrac{ 8\pm 0 }{2} \\ \mathbf{k} & \mathbf{=} & \mathbf{4} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline 16x^3-32x+16 : (x-1) = 16x^2+16x-16 \\ \hline \end{array}\)

 

 

laugh

heureka Apr 30, 2018

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