Let \(f(x)=6x^3+3x^2-5x+2\).

By the remainder theorem, the remainder is \(f(1)\).

\(f(1)=6(1)^3+3(1)^2-5(1)+2\)

\(=6+3-5+2\)

\(=6\)

So the remainder is \(6\).

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

By the remainder theorem, the remainder is \(f(-4)\).

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

\(=-64+16+44+12\)

\(=8\)

So the remainder is \(8\).

Let \(f(x)=x^4+2x^3-5x^2-2x+8\).

By the remainder theorem, the remainder is \(f(-1)\).

\(f(-1)=(-1)^4+2(-1)^3-5(-1)^2-2(-1)+8\)

\(=1-2-5+2+8\)

\(=4\)

So the remainder is \(4\).

Let \(f(x)=4x^3-x^2-18x+1\).

Since \(2x-1=0\) when \(x=\tfrac12\), the remainder is \(f\!\left(\tfrac12\right)\).

\(f\!\left(\tfrac12\right)=4\left(\tfrac12\right)^3-\left(\tfrac12\right)^2-18\left(\tfrac12\right)+1\)

\(=\tfrac12-\tfrac14-9+1\)

\(=\tfrac14-8\)

\(=-\tfrac{31}{4}\)

So the remainder is \(-\tfrac{31}{4}\).

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

By the remainder theorem, \(f(-2)=-37\).

\((-2)^3-3(-2)^2+a(-2)-7=-37\)

\(-8-12-2a-7=-37\)

\(-27-2a=-37\)

\(-2a=-10\)

\(a=5\)

Let \(f(x)=9x^3+bx-5\).

Since \(3x+2=0\) when \(x=-\tfrac23\), the remainder theorem gives \(f\!\left(-\tfrac23\right)=-13\).

\(9\left(-\tfrac23\right)^3+b\left(-\tfrac23\right)-5=-13\)

\(-\tfrac83-\tfrac{2b}{3}-5=-13\)

Multiply by \(3\):

\(-8-2b-15=-39\)

\(-23-2b=-39\)

\(-2b=-16\)

\(b=8\)

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

\(1+a+b-5=0\)

\(a+b=4\)   [1]

When divided by \(x+1\), the remainder is \(-6\), so \(f(-1)=-6\).

\((-1)^3+a(-1)^2+b(-1)-5=-6\)

\(-1+a-b-5=-6\)

\(a-b=0\)   [2]

Add [1] and [2]:

\(2a=4\)

\(a=2\)

Then \(b=2\).

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

\(3(-2)^3+a(-2)^2+b(-2)+8=0\)

\(-24+4a-2b+8=0\)

\(4a-2b=16\)

\(2a-b=8\)   [1]

When divided by \(x-1\), the remainder is \(15\), so \(f(1)=15\).

\(3+a+b+8=15\)

\(a+b=4\)   [2]

Add [1] and [2]:

\(3a=12\)

\(a=4\)

Then \(b=0\).

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

\(a\left(-\tfrac12\right)^3+7\left(-\tfrac12\right)^2+b\left(-\tfrac12\right)-8=0\)

\(-\tfrac a8+\tfrac74-\tfrac b2-8=0\)

Multiply by \(8\):

\(-a+14-4b-64=0\)

\(a+4b=-50\)   [1]

The remainder on division by \(x+1\) is \(7\), so \(f(-1)=7\).

\(-a+7-b-8=7\)

\(-a-b=8\)

\(a+b=-8\)   [2]

Subtract [2] from [1]:

\(3b=-42\)

\(b=-14\)

Then \(a=6\).

From part a, \(a=6\) and \(b=-14\).

So \(f(x)=6x^3+7x^2-14x-8\).

Since \(2x+1\) is a factor, divide by \(2x+1\):

\(6x^3+7x^2-14x-8=(2x+1)(3x^2+2x-8)\).

Now factor the quadratic:

\(3x^2+2x-8=(3x-4)(x+2)\).

Hence

\(f(x)=(2x+1)(3x-4)(x+2)\).

Let \(f(x)=x^3+6x^2+px-3\).

Since the remainder on division by \(x+1\) is \(R\), \(f(-1)=R\).

\((-1)^3+6(-1)^2+p(-1)-3=R\)

\(-1+6-p-3=R\)

\(2-p=R\)

So \(p=2-R\).   [1]

Since the remainder on division by \(x-3\) is \(-10R\), \(f(3)=-10R\).

\(27+54+3p-3=-10R\)

\(78+3p=-10R\)   [2]

Substitute \(p=2-R\) into [2]:

\(78+3(2-R)=-10R\)

\(84-3R=-10R\)

\(7R=-84\)

\(R=-12\)

Then \(p=2-(-12)=14\).

From part a, \(p=14\).

So the polynomial is \(f(x)=x^3+6x^2+14x-3\).

By the remainder theorem, the remainder on division by \(x-2\) is \(f(2)\).

\(f(2)=2^3+6(2)^2+14(2)-3\)

\(=8+24+28-3\)

\(=57\)

So the remainder is \(57\).

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

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

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

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

\(2a+b=-5\)   [1]

When divided by \(x+1\), the remainder is \(21\), so \(f(-1)=21\).

\((-1)^3+a(-1)^2+b(-1)+2=21\)

\(-1+a-b+2=21\)

\(a-b=20\)   [2]

Add [1] and [2]:

\(3a=15\)

\(a=5\)

Then \(b=-15\).

From part a, \(a=5\) and \(b=-15\).

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

Since \(x-2\) is a factor, divide by \(x-2\):

\(x^3+5x^2-15x+2=(x-2)(x^2+7x-1)\).

So \(f(x)=0\) gives

\((x-2)(x^2+7x-1)=0\).

One root is \(x=2\).

For the quadratic, use the formula:

\(x=\dfrac{-7\pm\sqrt{7^2-4(1)(-1)}}{2}\)

\(=\dfrac{-7\pm\sqrt{49+4}}{2}\)

\(=\dfrac{-7\pm\sqrt{53}}{2}\)

So the exact roots are \(2,\ \dfrac{-7-\sqrt{53}}{2},\ \dfrac{-7+\sqrt{53}}{2}\).