Answer: (a) \(a=6,\ b=1\); (b) \(\mathrm P(x)=(2x+1)(3x^2-x+2)\), and the quadratic factor has negative discriminant.

Use the remainder theorem: when a polynomial \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\). If \(x-a\) is a factor, then \(P(a)=0\).

(a) Since \(2x+1\) is a factor, \(x=-\frac12\) is a root.

So

\(\mathrm P\left(-\frac12\right)=0.\)

This gives

\(-\frac{a}{8}+\frac{b}{4}-\frac32+2=0.\)

Hence

\(-a+2b+4=0.\)

The remainder when divided by \(x+1\) is \(\mathrm P(-1)\), so

\(\mathrm P(-1)=-6.\)

Therefore

\(-a+b-3+2=-6,\)

which gives

\(-a+b+5=0.\)

Solving

\(-a+2b+4=0,\quad -a+b+5=0\)

gives

\(b=1,\quad a=6.\)

(b) With these values,

\(\mathrm P(x)=6x^3+x^2+3x+2.\)

Since \(2x+1\) is a factor, divide by \(2x+1\):

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

The quadratic factor has discriminant

\((-1)^2-4(3)(2)=1-24=-23.\)

Since this is negative, \(3x^2-x+2\) has no real roots.

The only real root therefore comes from \(2x+1=0\), so \(\mathrm P(x)=0\) has only one real root.