The zeroes of the function are the roots of the given
equationy the remainder theorem.
since \(x-1\) divides \(f(x)\).
then by long division
\(\frac{-x^{2}+x}{-6 x+6}\)
\(-6 x+6\)
\(\Longrightarrow 2 x^{2}-x^{2}-5 x^{2}+6\)
\(\equiv(x-1)\left(2 x^{2}-x-6\right)\)
we then factorise
\(2 x^{2}-x-6\) to obtain
\((x-2)(2 x+3)\)
\(\therefore\) The zeroes are
\(x=1 . x=2\) and \(x=-3 / 2\)