Optimization-factoring $\le_p$ Decision-factoring
You need to get an algorithm polynomial in $n$ where $n = \log N$ is the size of your input, i.e. the binary representation of $N$ (naive factorization algorithms are $O(N^2)$ which is of course exponential).

Given dec(k, x) solving the decision factoring problem, let's write a procedure to find the smallest prime divisor of $x$ using a dichotomy to keep logarithmic the number of steps.

sp(x):
  min = 2
  max = x - 1
  while (max &gt;= 1 + min):
    k = floor ((min + max) / 2)
    if dec(k, x):
      max = k
    else:
      min = k + 1
  if x % (min + 1) == 0:
    return (min + 1)
  else if x % min == 0:
    return min
  else:
    return 1


The procedure above calls dec $O(\log x)$ times and find the smallest prime number that divides $x$, or 1 if it does not exist. Now we can easily define the following factoring function:

factor(x):
  p = sp(x)
  if (p == 1):
    print x
  else:
    print p
    factor(x / p)


Factor is called less than $\log x$ times (since $p ≥ 2$) so the number of calls to dec is bounded by $\log^2 x$ times on values less than $x$. If the complexity of dec is polynomial, let's say bounded by an increasing $P(n)$, then the complexity of factor is still polynomial, bounded by $n^2P(n)$.