Let \(f(x)=x^4-3x^3-4x^2+5x+5\).

If \(x+1\) is a factor, then \(f(-1)=0\).

\(f(-1)=(-1)^4-3(-1)^3-4(-1)^2+5(-1)+5\)

\(=1+3-4-5+5\)

\(=0\)

So \(x+1\) is a factor by the factor theorem.

Let \(f(x)=2x^3-7x^2+9x-10\).

If \(2x-5\) is a factor, then \(f\!\left(\tfrac52\right)=0\).

\(f\!\left(\tfrac52\right)=2\left(\tfrac52\right)^3-7\left(\tfrac52\right)^2+9\left(\tfrac52\right)-10\)

\(=2\cdot\tfrac{125}{8}-7\cdot\tfrac{25}{4}+\tfrac{45}{2}-10\)

\(=\tfrac{250}{8}-\tfrac{350}{8}+\tfrac{180}{8}-\tfrac{80}{8}\)

\(=0\)

So \(2x-5\) is a factor by the factor theorem.

Let \(f(x)=x^3+ax^2-29x+12\).

Since \(x+4\) is a factor, \(f(-4)=0\).

\((-4)^3+a(-4)^2-29(-4)+12=0\)

\(-64+16a+116+12=0\)

\(16a+64=0\)

\(16a=-64\)

\(a=-4\)

Let \(f(x)=x^3+ax^2+bx-30\).

Since \(x-3\) is a factor, \(f(3)=0\).

\(3^3+a\cdot3^2+b\cdot3-30=0\)

\(27+9a+3b-30=0\)

\(9a+3b-3=0\)

\(9a=3-3b\)

\(a=\dfrac{1-b}{3}\)

Let \(f(x)=2x^3-x^2+ax+b\).

First factorise \(2x^2+x-1\):

\(2x^2+x-1=(2x-1)(x+1)\).

So \(2x-1\) and \(x+1\) are factors of \(f(x)\).

Since \(2x-1\) is a factor, \(f\!\left(\tfrac12\right)=0\):

\(2\left(\tfrac12\right)^3-\left(\tfrac12\right)^2+a\left(\tfrac12\right)+b=0\)

\(\tfrac14-\tfrac14+\tfrac a2+b=0\)

\(\tfrac a2+b=0\)

\(a=-2b\)

Since \(x+1\) is a factor, \(f(-1)=0\):

\(2(-1)^3-(-1)^2+a(-1)+b=0\)

\(-2-1-a+b=0\)

\(b=a+3\)

Substitute \(a=-2b\):

\(b=-2b+3\)

\(3b=3\)

\(b=1\)

Then \(a=-2\).

Let \(f(x)=2x^3+px^2+(2q-1)x+q\).

Since \(x-3\) is a factor, \(f(3)=0\):

\(2(3)^3+p(3)^2+(2q-1)(3)+q=0\)

\(54+9p+6q-3+q=0\)

\(9p+7q=-51\)

Since \(2x+1\) is a factor, \(f\!\left(-\tfrac12\right)=0\):

\(2\left(-\tfrac12\right)^3+p\left(-\tfrac12\right)^2+(2q-1)\left(-\tfrac12\right)+q=0\)

\(-\tfrac14+\tfrac p4-q+\tfrac12+q=0\)

\(\tfrac p4+\tfrac14=0\)

\(p=-1\)

Substitute into \(9p+7q=-51\):

\(9(-1)+7q=-51\)

\(7q=-42\)

\(q=-6\)

Substitute \(p=-1\) and \(q=-6\) into the expression:

\(f(x)=2x^3-x^2-13x-6\).

Now evaluate \(f(-2)\):

\(f(-2)=2(-2)^3-(-2)^2-13(-2)-6\)

\(=-16-4+26-6\)

\(=0\)

Since \(f(-2)=0\), \(x+2\) is a factor by the factor theorem.

Let \(f(x)=x^3+4x^2+7ax+4a\).

If \(x+a\) is a factor, then \(f(-a)=0\).

\((-a)^3+4(-a)^2+7a(-a)+4a=0\)

\(-a^3+4a^2-7a^2+4a=0\)

\(-a^3-3a^2+4a=0\)

\(-a(a^2+3a-4)=0\)

So \(a=0\) or \(a^2+3a-4=0\).

\((a+4)(a-1)=0\)

Hence \(a=0,-4,1\).

Let \(f(x)=x^3+px+q\) and \(g(x)=x^3+(1-p)x^2+19x-2q\).

Since \(x+1\) is a factor of \(f(x)\), \(f(-1)=0\):

\((-1)^3+p(-1)+q=0\)

\(-1-p+q=0\)

\(q=p+1\)

Since \(x+1\) is also a factor of \(g(x)\), \(g(-1)=0\):

\((-1)^3+(1-p)(-1)^2+19(-1)-2q=0\)

\(-1+1-p-19-2q=0\)

\(p+2q=-19\)

Substitute \(q=p+1\):

\(p+2(p+1)=-19\)

\(3p=-21\)

\(p=-7\)

Then \(q=-6\).

From part a, \(p=-7\) and \(q=-6\).

So the cubic is \(x^3-7x-6\).

Since \(x+1\) is a factor, divide by \(x+1\) to get \(x^2-x-6\).

Then \(x^2-x-6=(x-3)(x+2)\).

So \(x^3-7x-6=(x+1)(x-3)(x+2)\).

From part a, \(p=-7\) and \(q=-6\).

So the cubic is \(x^3+8x^2+19x+12\).

Since \(x+1\) is a factor, divide by \(x+1\) to get \(x^2+7x+12\).

Then \(x^2+7x+12=(x+3)(x+4)\).

So \(x^3+8x^2+19x+12=(x+1)(x+3)(x+4)\).

Let \(f(x)=x^4-x^3+px^2-11x+q\).

Since \(x-1\) is a factor, \(f(1)=0\):

\(1-1+p-11+q=0\)

\(p+q=11\)

Since \(x+2\) is a factor, \(f(-2)=0\):

\((-2)^4-(-2)^3+p(-2)^2-11(-2)+q=0\)

\(16+8+4p+22+q=0\)

\(4p+q=-46\)

Subtract the equations:

\((4p+q)-(p+q)=-46-11\)

\(3p=-57\)

\(p=-19\)

Then \(q=30\).

From part a, \(p=-19\) and \(q=30\), so

\(f(x)=x^4-x^3-19x^2-11x+30\).

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

\(f(x)=(x-1)(x^3-19x-30)\).

Since \(x+2\) is also a factor, divide \(x^3-19x-30\) by \(x+2\):

\(x^3-19x-30=(x+2)(x^2-2x-15)\).

Now factor the quadratic:

\(x^2-2x-15=(x-5)(x+3)\).

So \(f(x)=(x-1)(x+2)(x-5)(x+3)\).

Let \(f(x)=x^3-5x^2-4x+20\).

Try factors of \(20\). Since \(f(2)=0\), \(x-2\) is a factor.

Dividing gives \(x^2-3x-10\).

Now \(x^2-3x-10=(x-5)(x+2)\).

So \(x^3-5x^2-4x+20=(x-2)(x-5)(x+2)\).

The solutions are \(-2,2,5\).

Let \(f(x)=x^3+5x^2-17x-21\).

Try factors of \(-21\). Since \(f(-1)=0\), \(x+1\) is a factor.

Dividing gives \(x^2+4x-21\).

Now \(x^2+4x-21=(x+7)(x-3)\).

So \(x^3+5x^2-17x-21=(x+1)(x+7)(x-3)\).

The solutions are \(-7,-1,3\).

Let \(f(x)=2x^3-5x^2-13x+30\).

Try factors of \(30\). Since \(f(2)=0\), \(x-2\) is a factor.

Dividing gives \(2x^2-x-15\).

Now \(2x^2-x-15=(2x+5)(x-3)\).

So \(2x^3-5x^2-13x+30=(x-2)(2x+5)(x-3)\).

The solutions are \(-\tfrac52,2,3\).

Let \(f(x)=3x^3+17x^2+18x-8\).

Try factors of \(-8\). Since \(f(-2)=0\), \(x+2\) is a factor.

Dividing gives \(3x^2+11x-4\).

Now \(3x^2+11x-4=(3x-1)(x+4)\).

So \(3x^3+17x^2+18x-8=(x+2)(3x-1)(x+4)\).

The solutions are \(-4,-2,\tfrac13\).

Let \(f(x)=x^4+2x^3-7x^2-8x+12\).

Try factors of \(12\). Since \(f(1)=0\), \(x-1\) is a factor.

Dividing gives \(x^3+3x^2-4x-12\).

Now try factors of \(-12\) for the cubic. Since \(g(2)=0\), \(x-2\) is a factor.

Then \(x^3+3x^2-4x-12=(x-2)(x^2+5x+6)\).

And \(x^2+5x+6=(x+2)(x+3)\).

So \(x^4+2x^3-7x^2-8x+12=(x-1)(x-2)(x+2)(x+3)\).

The solutions are \(-3,-2,1,2\).

Let \(f(x)=2x^4-11x^3+12x^2+x-4\).

Try factors of \(-4\). Since \(f(1)=0\), \(x-1\) is a factor.

Dividing gives \(2x^3-9x^2+3x+4\).

Again \(g(1)=0\), so \(x-1\) is a factor of the cubic as well.

Dividing gives \(2x^2-7x-4\).

Now \(2x^2-7x-4=(2x+1)(x-4)\).

So \(2x^4-11x^3+12x^2+x-4=(x-1)^2(2x+1)(x-4)\).

The distinct solutions are \(-\tfrac12,1,4\). The root \(x=1\) is repeated.

Let the three roots be \(e-f\), \(e\), and \(e+f\).

Then

\(x^3+ax^2+bx+c=(x-(e-f))(x-e)(x-(e+f))\).

This is

\((x-e+f)(x-e)(x-e-f)\)

\(=(x-e)\bigl((x-e)^2-f^2\bigr)\)

\(=(x-e)(x^2-2ex+e^2-f^2)\)

\(=x^3-3ex^2+(3e^2-f^2)x-e^3+ef^2\).

Comparing coefficients gives

\(a=-3e\), \(b=3e^2-f^2\), \(c=-e^3+ef^2\).

Now

\(eb=3e^3-ef^2\)

so

\(eb+c=2e^3\).

Also \(a^3=(-3e)^3=-27e^3\).

Therefore

\(2a^3=2(-27e^3)=-54e^3=-27(2e^3)=-27(eb+c)\).

So

\(2a^3+27c=-27eb=9ab\).

Consider \(y=3x^4+4x^3-12x^2+k\).

For the equation to have four real roots, the graph must cross the \(x\)-axis four times. So the two outer turning points must lie below the axis and the middle turning point must lie above it.

Differentiate:

\(\dfrac{dy}{dx}=12x^3+12x^2-24x\)

\(=12x(x^2+x-2)\)

\(=12x(x-1)(x+2)\).

So the turning points occur at \(x=-2,0,1\).

Now find the corresponding \(y\)-values:

At \(x=-2\): \(y=3(16)+4(-8)-12(4)+k=k-32\).

At \(x=0\): \(y=k\).

At \(x=1\): \(y=3+4-12+k=k-5\).

For four real roots, we need

\(k>0\),

\(k-5<0\),

and \(k-32<0\).

These give \(k>0\) and \(k<5\). The condition \(k<32\) is already satisfied.

So the required set of values is \(0<k<5\).