Given that:
(x-1) is a factor of k²x⁴-3kx²+2
Solving for k
To solve for k, we are going to equate the factor to zero and solve for x. After that, we substitute the value of x it into the equation , k²x⁴-3kx²+2=0 and solve for k.
Therefore,
![\begin{gathered} x-1=0 \\ x=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/xzovlzhh66fbtyj7sg6kifanquqr7ao1do.png)
Substitute x = 1, into the equation and solve for k
![\begin{gathered} k^2x^4-3kx^2+2=0 \\ \text{where,} \\ x=1 \\ k^2(1)^4-3k(1)^2+2=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/24v456n9v8kcvfymmravi3zrxe4e204fmx.png)
Simplifying the equation above
![\begin{gathered} k^2(1)-3k(1)+2=0 \\ k^2-3k+2=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/n6jdagzbfpso5r5p5vcv5u8x658jqf6686.png)
Factorizing the equation
![\begin{gathered} k^2-1k-2k+2=0 \\ k(k-1)-2(k-1)=0 \\ (k-1)(k-2)=0 \\ k-1=0\text{ or k-2=0} \\ k=1\text{ or k=2} \\ k=1\text{ or 2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/272nw5qz29o07dr26wohz3hw5olwhs34yt.png)
Hence, the possible values of k are,
![k=1\text{ or 2}](https://img.qammunity.org/2023/formulas/mathematics/college/ipi7jcvj5eo8l543xszgogqil234yzvf4d.png)