Questions & Answers for: common factors

...common factors of the numerator and the denominator of the following rational expressions. 1. 4c/6c 2. 2xy/3x 3. 8a2c/-4ac2 4. -14abc/12ac 5. (-3r)(-4s)/-18r2 1.2c 2.x 3.8ac 4 .2ac 5.12r...hope these ...

...http://www.mathsisfun.com/greatest-common-factor.html To find the common factors, you have to list all factors of both numbers: 96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96 124: 1, 2, 4, 31, 62, ...

...common divisor (or HCF) but that should not be difficult to do by hand. Decompose the first number in prime factors. If a prime factor is repeated use the exponent notation: That helps Decompose the ...

common. For 84, the factors are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84. For 42, the factors are: 1, 2, 3, 6, 7, 14, 21, and 42. The largest factor that they have in common is 42, so that is the

...common factors? Do a prime factorization of the two numbers. Use condensed (power) notation for each prime factor. Example N1=(2^5)*(3^4)*(7^2)(11^3) N2=(2^4)*(5^6)*(11) 2 is a factor in N1 and N2, ...

common factors. So let's factor 711 and 22. 1 x 711 9 x 79 3 x 3 x 79 I don't think I can factor 79. We should check the numbers from 2 up to the square root of 79, or approximately 9. Factoring 22, 1

...7, 8, 14, 28 and 56. The factors of 49 are 1, 7, 49. What is the least common factor of both numbers? 1 So the least common factor of any two numbers is 1. Any questions, let me know. Good luck, ...

...common factor of 14, 42, and 56. Factors of 14 : 1, 2, 7, 14 Factors of 42 : 1, 2, 3, 6, 7, 14, 21, 42 Factors of 56 : 1, 2, 4, 7, 8, 14, 28, 56 What is the highest factor, in all three (which makes ...

...common factor is 11. 77, 132, 143 Factors of 77 1 7 11 77 Factors of 132 1 2 3 4 6 11 12 22 33 44 66 132 Factors of 143 1 11 13 143 The highest (greatest) common factor, the highest number in each of ...

...factors of 3 are 3 and 1, this must be the greatest common factor of this component. Onto the next component, what is the greatest common factor of k^2 and k. The answer is k, because k^2 doesn't go ...