Answer: \(a=-27\), \(b=3\); \(\mathrm p(x)=(2x+1)(5x-1)(x-3)\); remainder \(3\).

Since \(\mathrm p(x)\) is divisible by \(2x+1\), the factor theorem gives \(\mathrm p\left(-\frac12\right)=0\).

So

\(-\frac54+\frac a4+5+b=0.\)

Multiplying by \(4\),

\(a+4b=-15. \tag{1}\)

The remainder when \(\mathrm p(x)\) is divided by \(x+1\) is \(\mathrm p(-1)\). Hence

\(\mathrm p(-1)=-24.\)

So

\(10(-1)^3+a(-1)^2-10(-1)+b=-24,\)

which gives

\(a+b=-24. \tag{2}\)

Subtracting (2) from (1),

\(3b=9,\)

so \(b=3\). Then \(a=-27\).

Thus

\(\mathrm p(x)=10x^3-27x^2-10x+3.\)

Since \(2x+1\) is a factor, division gives

\(\mathrm p(x)=(2x+1)(5x^2-16x+3).\)

Factorising the quadratic,

\(5x^2-16x+3=(5x-1)(x-3).\)

Therefore

\(\mathrm p(x)=(2x+1)(5x-1)(x-3).\)

The remainder when \(\mathrm p(x)\) is divided by \(x\) is \(\mathrm p(0)\), so the remainder is \(3\).