Answer: \(a=7\), \(b=14\); \(\mathrm{p}(x)=0\) has only one real root.

(a) Since \(x+2\) is a factor of \(\mathrm p(x)\), the factor theorem gives

\(\mathrm p(-2)=0.\)

Substitute \(x=-2\) into \(\mathrm p(x)=2x^3+ax^2+13x+b\):

\(2(-2)^3+a(-2)^2+13(-2)+b=0.\)

So

\(-16+4a-26+b=0,\)

which gives

\(4a+b=42. \tag{1}\)

The remainder when \(\mathrm p(x)\) is divided by \(x+1\) is \(\mathrm p(-1)\). This remainder is \(6\), so

\(2(-1)^3+a(-1)^2+13(-1)+b=6.\)

Hence

\(-2+a-13+b=6,\)

so

\(a+b=21. \tag{2}\)

Subtracting (2) from (1),

\(3a=21,\)

so \(a=7\). Then from \(a+b=21\),

\(b=14.\)

(b) Therefore

\(\mathrm p(x)=2x^3+7x^2+13x+14.\)

Since \(x+2\) is a factor, divide by \(x+2\):

\(\mathrm p(x)=(x+2)(2x^2+3x+7).\)

Thus \(x=-2\) is one real root. The other possible roots come from

\(2x^2+3x+7=0.\)

The discriminant is

\(3^2-4(2)(7)=9-56=-47.\)

Since this is negative, the quadratic has no real roots. Hence \(\mathrm p(x)=0\) has only one real root.