Math Problem Solver

The prime factorization of \(56\) is \(2^3 \cdot 7\).

 The prime factorization of an integer is the decomposition of that integer into the product of powers of prime numbers. Begin by listing the first ten primes: $2,3,5,7,11,13,17,19,23,29$.

 Divide the integer by the smallest divisible prime. Then divide the result by its smallest divisible prime. Continue until the result is prime, and multiply the primes together to get the prime factorization.

$$\begin{aligned}2&\underline{\big | 56}\\2&\underline{\big | 28}\\2&\underline{\big | 14}\\& \: \: 7\end{aligned}$$

$56 = 2 \cdot 2 \cdot 2 \cdot 7 = 2^3 \cdot 7$