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 \(a=-2\).