Answer :

k can be (-1/3) or (2/3).
since (x-3) is a factor. if you put x=3 in the equation, it must be equal to zero. So the equation becomes (9k2 - 3k -2 = 0). if you solve this you will get k=(-1/3) and k=(2/3). You can substitute this k in the original equation and check, (x-3) is a factor.
[tex]w(x)=k^2x^2-kx-2\\\\(x-3)\ is\ a\ factor\ w(x)\ then\ w(3)=0.\\\\Substitute:\\\\k^2\cdot3^2-k\cdot3-2=0\\9k^2-3k-2=0\\9k^2-6k+3k-2=0\\3k(3k-2)+1(3k-2)=0\\(3k-2)(3k+1)=0\iff3k-2=0\ \vee\ 3k+1=0\\3k=2\ \vee\ 3k=-1\\\\\underline{\underline{k=\frac{2}{3}\ \vee\ k=-\frac{1}{3}}}[/tex]

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