Answer: normal \(y=-x+5\); other \(x\)-coordinates \(-1-\sqrt2\) and \(-1+\sqrt2\).

(a) Differentiate

\(y=x^3+x^2-4x+6.\)

This gives

\(\frac{dy}{dx}=3x^2+2x-4.\)

At \(x=1\),

\(\frac{dy}{dx}=3+2-4=1.\)

So the gradient of the normal is \(-1\). The normal passes through \((1,4)\), so

\(y-4=-(x-1).\)

Hence

\(y=-x+5.\)

(b) To find where the normal meets the curve, solve

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

Rearrange:

\(x^3+x^2-3x+1=0.\)

Since \(x=1\) is the original point of contact, \(x-1\) is a factor. Dividing gives

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

The other two intersections are found from

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

Using the quadratic formula,

\(x=\frac{-2\pm\sqrt{4+4}}{2}.\)

Therefore

\(x=-1\pm\sqrt2.\)

The two \(x\)-coordinates where the normal cuts the curve again are

\(-1-\sqrt2\quad\text{and}\quad -1+\sqrt2.\)