How do you write the prime factorization of 48?

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Answer 1 · 2018-03-16T07:13:47

$48 = 2 \times 2 \times 2 \times 2 \times 3 = 2^{4} \times 3$ $48 = 2xx2xx2xx2xx3 = 2^4xx3$

Expand $48$ $48$ into the product of all its prime factors:

$48 = 8 \times 6$ $48 = 8xx6$

$48 = 2 \times 4 \times 2 \times 3$ $48 = 2xx4xx2xx3$

$48 = 2 \times 2 \times 2 \times 2 \times 3$ $48 = 2xx2xx2xx2xx3$

(It does not matter which factors you start with, you will always get to the same answer.)

Or Use division, and divide by only prime factors.

$2|ul48$
$2|ul24$
$2|ul12$
$2|ul6$
$3|ul3$
$" "|ul1" "larr$ divide till you get to $1$

$:. 48 = 2xx2xx2xx2xx3 = 2^4xx3$