(a) The quadratic quotient has leading coefficient 2 and constant term \(-3\), so write

\[p(x)=(x^2+x+1)(2x^2+cx-3).\]

Expanding gives \(2x^4+(c+2)x^3+(c-1)x^2+(c-3)x-3\). Comparing the \(x^2\) coefficient gives \(c-1=4\), so \(c=5\). Thus \(\boxed{a=7,\ b=2}\).

(b) Factorise the quotient: \[2x^2+5x-3=(2x-1)(x+3).\]

Hence \(p(x)=(x^2+x+1)(2x-1)(x+3)\), so \(x+3\) is a factor, as required.