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# Algebra 2

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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

#1
+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.

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

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

Apr 30, 2018

#1
+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.

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

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

heureka Apr 30, 2018