Answer: \(x\)-intercepts \(1,\frac52,5\), \(y\)-intercept \(25\); \(1\leqslant x\leqslant\frac52\) or \(x\geqslant5\).

(a) The \(x\)-intercepts occur when

\((1-x)(x-5)(2x-5)=0.\)

So the roots are

\(1-x=0,\qquad x-5=0,\qquad 2x-5=0.\)

Hence the \(x\)-intercepts are

\(x=1,\quad x=5,\quad x=\dfrac52.\)

The corresponding points are

\((1,0),\quad \left(\dfrac52,0\right),\quad (5,0).\)

The \(y\)-intercept is found by putting \(x=0\):

\(y=(1)(-5)(-5)=25.\)

So the \(y\)-intercept is \((0,25)\).

The leading term of the cubic is

\(( -x)(x)(2x)=-2x^3.\)

Since the leading coefficient is negative, the graph rises on the left and falls on the right. Each root is simple, so the graph crosses the \(x\)-axis at each intercept.

(b) The inequality asks where the cubic is less than or equal to zero. From the sketch, the graph is below or on the \(x\)-axis between the first two roots, and to the right of the third root.

Therefore

\(1\leqslant x\leqslant\dfrac52\quad\text{or}\quad x\geqslant5.\)