Intermediate Algebra
Factors: 2⋅2⋅2⋅5⋅52\(\cdot\)2\(\cdot\)2\(\cdot\)5\(\cdot\)5. Extract pairs (22 and 55) for 1010 outside; leave the unpaired 22 inside.
Factors: 2⋅10⋅102\(\cdot\)10\(\cdot\)10. Extract the pair of 1010s outside; leave the 22 inside.
Factors: 2⋅2⋅5⋅52\(\cdot\)2\(\cdot\)5\(\cdot\)5. Multiply to get 100100 (1010 outside), then add a 22 inside to balance.
Factors: 2⋅2⋅2⋅5⋅52\(\cdot\)2\(\cdot\)2\(\cdot\)5\(\cdot\)5. Divide the prime factors by two to get 1010 outside and 22 inside.