Question

The number of zeros at the end of (2^123- 2^122 -2^121)(3^223-3^222-3^221) is​

Answer 1

answer is attached to the file

thank you

Answer 2

Answer:

One zero .

Step-by-step explanation:

(2^(123) - 2^(122) -2^(121) )( 3^(223) -3^(222) -3^(221))

The above equation can be factored by taking commons.

= 2^(121) [ 2^2 - 2 - 1 ] × 3^(121) [ 3^2 - 3 - 1 ]

= 2^(121) × 3^(121) [ 4 - 2 - 1 ] × 3^(121) [ 9 - 4 ]

= 2^(121) × 3^(121) × 3 × 5

Since there is only one 5 in the given equation.

There will only be one zero.