Answer: (a) \(f(x)=-3(3x+1)(x-1)(2x-5)\); (b) \(-\frac13\lt x\lt 1\) or \(x\gt \frac52\).

Work through the problem step by step, using exact values where possible before giving any final numerical approximation.

(a) From the graph, the roots are \(x=-\frac13\), \(x=1\) and \(x=\frac52\).

So \(f(x)\) has factors \((3x+1)\), \((x-1)\) and \((2x-5)\).

Hence

\(f(x)=k(3x+1)(x-1)(2x-5)\).

From the graph, the \(y\)-intercept is \(-15\). Therefore \(f(0)=-15\).

Substitute \(x=0\):

\(f(0)=k(1)(-1)(-5)=5k\).

So \(5k=-15\), giving \(k=-3\).

Therefore

\(f(x)=-3(3x+1)(x-1)(2x-5)\).

(b) From the graph, \(f(x)\lt 0\) where the curve lies below the \(x\)-axis.

This occurs between \(-\frac13\) and \(1\), and again to the right of \(\frac52\).

Thus

\(-\frac13\lt x\lt 1\quad\text{or}\quad x\gt \frac52\).