2 3 5 7 11 13 17
2 is already factored as low as it will go. (In other words, it’s one of those prime numbers on the list above. ) We’ll ignore this for now and try to divide 49 instead. 49 can’t be evenly divided by 2, or by 3, or by 5. You can test this yourself using a calculator or long division. Because these don’t give us nice, whole number results, we’ll ignore them and keep trying. 49 can be evenly divided by seven. 49 ÷ 7 = 7, so 49 = 7 x 7. Rewrite the problem: √(2 x 49) = √(2 x 7 x 7).
Even if it’s possible to keep factoring, you don’t need to once you’ve found two identical factors. For example, √(16) = √(4 x 4) = 4. If we kept on factoring, we’d end up with the same answer but have to do more work: √(16) = √(4 x 4) = √(2 x 2 x 2 x 2) = √(2 x 2)√(2 x 2) = 2 x 2 = 4.
√180 = √(2 x 90) √180 = √(2 x 2 x 45) √180 = 2√45, but this can still be simplified further. √180 = 2√(3 x 15) √180 = 2√(3 x 3 x 5) √180 = (2)(3√5) √180 = 6√5
70 = 35 x 2, so √70 = √(35 x 2) 35 = 7 x 5, so √(35 x 2) = √(7 x 5 x 2) All three of these numbers are prime, so they cannot be factored further. They’re all different, so there’s no way to “pull out” an integer. √70 cannot be simplified.
12 = 1 22 = 4 32 = 9 42 = 16 52 = 25 62 = 36 72 = 49 82 = 64 92 = 81 102 = 100
√1 = 1 √4 = 2 √9 = 3 √16 = 4 √25 = 5 √36 = 6 √49 = 7 √64 = 8 √81 = 9 √100 = 10
√50 = √(25 x 2) = 5√2. If the last two digits of a number end in 25, 50, or 75, you can always factor out 25. √1700 = √(100 x 17) = 10√17. If the last two digits end in 00, you can always factor out 100. √72 = √(9 x 8) = 3√8. Recognizing multiples of nine is often helpful. There’s a trick to it: if all digits in a number add up to nine, then nine is always a factor. √12 = √(4 x 3) = 2√3. There’s no special trick here, but it’s usually easy to check whether a small number is divisible by 4. Keep this in mind when looking for factors.
√72 = √(9 x 8) √72 = √(9 x 4 x 2) √72 = √(9) x √(4) x √(2) √72 = 3 x 2 x √2 √72 = 6√2
