If the error in the iterate is denoted by then Newton's method may be written

.

The Taylor expansions for the two functions are

and

If the zero is of order 1 then so when you divide (1) by (2) you get plus higher order terms.

That means that the 's on the RHS of the method cancel and you get second order convergence.

If though the zero is of order 2 then so that when the division is carried out the leading term is and in which case the 's no longer cancel out and you are left with first order convergence.

To restore the quadratic convergence the fractional term in the method is multiplied by 2.

Now just follow the reasoning 'up the line' to a order zero.