Answer: (a) \(f(x)=(x-1)(x-4)\) or \(f(x)=-(x-1)(x-4)\); (b) \(p(x)=5x^3+14x^2+7x-2\).

Answer: (a) \(f(x)=(x-1)(x-4)\) or \(f(x)=-(x-1)(x-4)\); (b) \(p(x)=5x^3+14x^2+7x-2\).

(a) From the graph of \(y=|f(x)|\), the zeros occur at \(x=1\) and \(x=4\). Hence \(f(x)\) must have the factors \((x-1)\) and \((x-4)\).

Also, when \(x=0\), the graph has value \(4\), so

\(|f(0)|=4.\)

If \(f(x)=k(x-1)(x-4)\), then

\(|f(0)|=|4k|=4,\)

so \(k=1\) or \(k=-1\). Therefore

\(f(x)=(x-1)(x-4) \quad\text{or}\quad f(x)=-(x-1)(x-4).\)

(b) Since the roots are \(\frac15\), \(n\) and \(n+1\), and the leading coefficient is \(5\),

\(p(x)=(5x-1)(x-n)(x-(n+1)).\)

For \(5x^3+ax^2+bx-2\), the product of the roots is

\(-\frac{-2}{5}=\frac25.\)

Thus

\(\frac15 n(n+1)=\frac25.\)

So

\(n(n+1)=2.\)

This gives

\(n^2+n-2=0,\)

or

\((n+2)(n-1)=0.\)

Since \(n\) is negative, \(n=-2\). Hence the roots are \(\frac15\), \(-2\) and \(-1\), so

\(p(x)=(5x-1)(x+2)(x+1).\)

Now

\((x+2)(x+1)=x^2+3x+2.\)

Therefore

\(p(x)=(5x-1)(x^2+3x+2)=5x^3+14x^2+7x-2.\)