Prime Factorization: Practice

A prime factorization breaks a number into the prime building blocks you multiply to rebuild it.

$12 = 2 \times 2 \times 3$

Every factor on the right is prime, so this is a product of primes only, not a list of all factors or a sum.

Split $6$ into a product of primes. Both $2$ and $3$ are prime.

$6 = 2 \times 3$

The form $1 \times 6$ uses $1$ (not prime) and $6$ (not prime), and a factorization is a product, not a sum.

Divide $10$ by the $2$ already shown to find its partner.

$10 \div 2 = 5$

Since $5$ is prime, the tree is finished: $10 = 2 \times 5$.

A prime has no factors except $1$ and itself, so it cannot be split further.

$13 = 13$

The others split: $9 = 3 \times 3$, $15 = 3 \times 5$, and $21 = 3 \times 7$.

Keep splitting until every factor is prime: $8 = 2 \times 4 = 2 \times 2 \times 2$.

$8 = 2 \times 2 \times 2 = 2^3$

The form $2 \times 4$ stops too early ($4$ is composite), and $4^2 = 16$, not $8$.

Divide $14$ by the $7$ already shown.

$14 \div 7 = 2$

Since $2$ is prime, the factorization is $14 = 2 \times 7$.

Split $15$ into primes. Both $3$ and $5$ are prime.

$15 = 3 \times 5$

The other choices keep a composite factor or use a sum instead of a product.

First expand the power, then multiply.

$2^2 \times 3 = (2 \times 2) \times 3 = 4 \times 3 = 12$

So $2^2 \times 3 = 12$.

Check each for a factor other than $1$ and itself.

$9 = 3 \times 3$

So $9$ is composite and never appears in a prime factorization, while $2$, $3$, and $5$ are all prime.

Split $25$ into primes: $25 = 5 \times 5$, then write the repeat as a power.

$25 = 5 \times 5 = 5^2$

The form $5 \times 10$ leaves $10$ composite, and $2^5 = 32$, not $25$.

Prime-power form lists each distinct prime once with an exponent for how many times it appears.

$2 \times 2 \times 2 \times 5 = 2^3 \times 5$

The expanded product and the forms using $8$ or $4$ are equal in value but not written as prime powers.

Split $20$ until only primes remain: $20 = 4 \times 5 = 2 \times 2 \times 5$.

$20 = 2 \times 2 \times 5 = 2^2 \times 5$

The forms $4 \times 5$ and $2 \times 10$ keep a composite factor, so they are not finished.