Answer: \(a=4\), \(\mathrm{p}(x)=2(2x+1)(4x^2+1)\), and the only real root is \(x=-\frac12\).

Use the factor theorem and the remainder theorem to turn the given information into equations for the unknown constants. After finding the constants, factorise the polynomial if required.

Since \(2x+1\) is a factor, \(x=-\frac12\) is a root. Therefore \(\mathrm{p}(-\frac12)=0\).

Substitute \(x=-\frac12\):

\(a^2\left(-\frac12\right)^3+2a\left(-\frac12\right)^2+a\left(-\frac12\right)+2=0.\)

This gives

\(-\frac{a^2}{8}+\frac{a}{2}-\frac{a}{2}+2=0,\)

so

\(-\frac{a^2}{8}+2=0.\)

Hence \(a^2=16\). Since \(a\) is positive, \(a=4\).

Now

\(\mathrm{p}(x)=16x^3+8x^2+4x+2.\)

Factorising gives

\(\mathrm{p}(x)=2(2x+1)(4x^2+1).\)

The factor \(4x^2+1\) has no real roots, since \(4x^2+1=0\) would require \(x^2=-\frac14\).

Therefore the only real root comes from \(2x+1=0\), namely \(x=-\frac12\).

The result has been obtained using the exact condition from the question, so no additional solutions are introduced.