A polynomial \(f(x)\) has a factor \((x-a)^2\). Show that \((x-a)\) is a factor of \(f'(x)\).

Since \(g(x)\) and \(g'(x)\) are polynomials, we have shown that \(f'(x)\) has \((x-a)\) as a factor.

We can understand this result graphically as follows. If \((x-a)^2\) is a factor, then \(f(x)=0\) has a repeated root at \(x=a\) and the graph of \(y=f(x)\) touches the \(x\)-axis here. The gradient of the graph must therefore be zero at this point, \(f'(a)=0\) and therefore \((x-a)\) is a factor of \(f'(x)\).

The equation \(x^3 - 4x^2 - 3x + p = 0\) has two equal roots; find the two possible values of \(p\).

Let \(f(x) = x^3 - 4x^2 - 3x + p\), and let’s assume the repeated root is at \(x = a\). Then \((x-a)^2\) is a factor of \(f(x)\).

From the result above, \((x-a)\) is a factor of \[\begin{equation*} f'(x) = 3x^2 - 8x - 3 = (3x + 1)(x - 3) \end{equation*}\]so that \(a = 3\) or \(a = -\dfrac{1}{3}\).

If \(a = 3\), then substituting \(x=a\) into \(x^3-4x^2-3x+p=0\), we have \(p = 18\).

If \(a = -\dfrac{1}{3}\), then from \(x^3-4x^2-3x+p=0,\) we have \(p = -\dfrac{14}{27}\).

Thus \(p = 18\) and \(p = -\dfrac{14}{27}\) are the two possible values for \(p\).