Answer: \(a=2\), \(b=0\).

Since \(x-3\) is a factor of \(\mathrm{p}(x)\), the factor theorem gives

\(\mathrm{p}(3)=0\).

Substitute \(x=3\) into

\(\mathrm{p}(x)=ax^3-7x^2-bx+9\):

\(27a-7(9)-3b+9=0\).

This simplifies to

\(27a-3b-54=0\).

Dividing by \(3\),

\(9a-b=18\). \(\quad\text{(1)}\)

When \(\mathrm{p}(x)\) is divided by \(x+2\), the remainder is \(\mathrm{p}(-2)\). The remainder is given as \(-35\), so

\(\mathrm{p}(-2)=-35\).

Substituting \(x=-2\),

\(-8a-7(4)+2b+9=-35\).

So

\(-8a+2b-19=-35\),

which gives

\(-8a+2b=-16\).

Dividing by \(2\),

\(-4a+b=-8\). \(\quad\text{(2)}\)

Add equations (1) and (2):

\((9a-b)+(-4a+b)=18-8\).

Thus

\(5a=10\),

so

\(a=2\).

Substitute into \(-4a+b=-8\):

\(-4(2)+b=-8\),

so

\(b=0\).

Answer: \(a=2\), \(b=-1\).

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

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

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

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

So

\(-8+4a-2b-2=0,\)

which simplifies to

\(4a-2b=10.\)

Divide by \(2\):

\(2a-b=5. \tag{1}\)

By the remainder theorem, the remainder when \(\mathrm p(x)\) is divided by \(x-3\) is \(\mathrm p(3)\). This remainder is \(40\), so

\(3^3+a(3)^2+b(3)-2=40.\)

Therefore

\(27+9a+3b-2=40,\)

so

\(9a+3b=15.\)

Divide by \(3\):

\(3a+b=5. \tag{2}\)

From (1),

\(b=2a-5.\)

Substitute this into (2):

\(3a+(2a-5)=5.\)

So

\(5a=10,\)

giving

\(a=2.\)

Then

\(b=2(2)-5=-1.\)

Therefore

\(a=2,\qquad b=-1.\)

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.