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I have no idea how to go about this. Any Hint?

Suppose that $(x-3)$ is a factor of

$$kx^3 - 6x^2 + 2kx - 12.$$

Solve for $k$.

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Hint: if $x-3$ is a factor, then when $x=3$ the expression evaluates to 0. So, plug in 3 for $x$ and see what you get...

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Hint:

According to the Remainder Theorem, for any polynomial $f(x)$, if $f(a)=0$, $(x-a) $ is a factor of $f(x)$.

So in your question, if $(x-3)$ is a factor, then for $a=3$, $f(x)=?$

With this fact, you can list a equation in terms of $k,x$ and also a constant("?"),where $x=a=3$.

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$$kx^3 - 6x^2 + 2kx - 12$$ can be factored as $$(kx-6)(x^2+2)$$Now, $(x-3)$ isn't obviously a factor of $(x^2+2)$, so try making it a factor of $(kx-6)$. That is, $$(kx-6) = a(x-3)$$What must $a$ be?

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