Answer: normal \(y-3=\frac1{11}(x-1)\); \(a=2\); \(x=-2\) or \(x=\frac32\).

(a) Differentiate

\(\mathrm f(x)=4x^3-4x^2-15x+18.\)

This gives

\(\mathrm f'(x)=12x^2-8x-15.\)

When \(x=1\),

\(\mathrm f(1)=4-4-15+18=3\)

and

\(\mathrm f'(1)=12-8-15=-11.\)

So the tangent gradient is \(-11\), and the normal gradient is

\(\frac1{11}.\)

The normal passes through \((1,3)\), so its equation is

\(y-3=\frac1{11}(x-1).\)

(b) Since \(x+a\) is a factor, a root is \(x=-a\). Test \(x=-2\):

\(\mathrm f(-2)=4(-8)-4(4)-15(-2)+18.\)

So

\(\mathrm f(-2)=-32-16+30+18=0.\)

Therefore \(x+2\) is a factor, so

\(a=2.\)

Divide \(4x^3-4x^2-15x+18\) by \(x+2\):

\(\mathrm f(x)=(x+2)(4x^2-12x+9).\)

The quadratic factor is a square:

\(4x^2-12x+9=(2x-3)^2.\)

Thus

\(\mathrm f(x)=(x+2)(2x-3)^2.\)

Hence

\(x=-2\quad\text{or}\quad 2x-3=0.\)

Therefore

\(x=-2\quad\text{or}\quad x=\frac32.\)