Answer: intercepts \((-3,0)\), \((1,0)\), \(\left(\dfrac52,0\right)\), \((0,-15)\); \(-3\leqslant x\leqslant1\) or \(x\geqslant\dfrac52\).
We start with the main method. Use the roots, intercepts and sign of the leading coefficient to determine the shape of the cubic and its modulus.
The \(x\)-intercepts are found by setting each factor equal to zero:
\(2x-5=0,\quad x+3=0,\quad 1-x=0.\)
Hence
\(x=\frac52,\quad x=-3,\quad x=1.\)
The \(y\)-intercept is found by putting \(x=0\):
\(y=(-5)(3)(1)=-15.\)
The leading term is \((2x)(x)(-x)=-2x^3\), so the cubic rises on the left and falls on the right.
For the inequality, use the sign of the cubic across its roots \(-3\), \(1\), and \(\frac52\).
Since the graph is at or below the \(x\)-axis on \([-3,1]\) and from \(\frac52\) onwards,
\((2x-5)(x+3)(1-x)\leqslant0\) gives
\(-3\leqslant x\leqslant1\quad\text{or}\quad x\geqslant\frac52.\)
For \(y=\left|(2x-5)(x+3)(1-x)\right|\), reflect the parts of the cubic below the \(x\)-axis above the \(x\)-axis, keeping the same \(x\)-intercepts.
This completes the solution and gives the required result.
