Answer: \(p(x)=(x+4)(2x^2+3x+10)\), so the only real root is \(x=-4\).

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) Substitute \(x=-4\):

\(p(-4)=2(-4)^3+11(-4)^2+22(-4)+40.\)

So

\(p(-4)=-128+176-88+40=0.\)

Therefore \(x=-4\) is a root.

(b) Since \(x=-4\) is a root, \(x+4\) is a factor of \(p(x)\). Dividing \(2x^3+11x^2+22x+40\) by \(x+4\) gives

\(p(x)=(x+4)(2x^2+3x+10).\)

For the quadratic factor, the discriminant is

\(b^2-4ac=3^2-4(2)(10)=9-80=-71.\)

Since the discriminant is negative, \(2x^2+3x+10=0\) has no real roots. Hence \(p(x)=0\) has no other real roots.