Answer: \(\mathrm p(x)=(4x+1)(3x-1)^2\), so \(x=\frac13\) is a repeated root.

(a) Substitute \(x=-0.25=-\frac14\):

This is

\(-\frac{36}{64}-\frac{15}{16}+\frac12+1=0.\)

So \(x=-0.25\) is a root.

(b) Since \(x=-\frac14\) is a root, \(4x+1\) is a factor.

Factorising the polynomial gives

\(36x^3-15x^2-2x+1=(4x+1)(9x^2-6x+1).\)

The quadratic factor is

\(9x^2-6x+1=(3x-1)^2.\)

Therefore

\(\mathrm p(x)=(4x+1)(3x-1)^2.\)

So \(x=\frac13\) is a repeated root of \(\mathrm p(x)=0\).