Question

How do you find the greatest common factor of \[140n,140{{m}^{2}},80{{m}^{2}}\]?

Answer 1

Hint: The greatest common factor is the factor that divides a given number of terms. In order to find the greatest common factor of these three monomials, we write these in factor form of the prime numbers and then find the factor that is common in all the terms.

Formula used:
In order to find the prime factors, we start by dividing the number by the least prime number \[2\] and divide until we get a remainder. After that, we divide it by \[3,5,7\] and so on until we are left with prime numbers only .Secondly, we write all the prime numbers in the multiplication form . For example for \[20\], we start by dividing \[20\] by\[2\] , and after dividing twice, we get \[5\] as remainder which is also a prime number so we write, \[\left( 2\times 2\times 5 \right)\].
And the product of prime numbers that are common form the Greatest Common Factor.

Complete step by step solution:
Firstly, in order to find the greatest common factor of \[140n,140{{m}^{2}},80{{m}^{2}}\], we write the prime factors of the all the terms:
For the first term, i.e.
\[\begin{align}
  & 140n \\
 & \Rightarrow 2\times 2\times 5\times 7\times n \\
\end{align}\]
For the second term, i.e.
\[\begin{align}
  & 140{{m}^{2}} \\
 & \Rightarrow 2\times 2\times 5\times 7\times {{m}^{2}} \\
\end{align}\]
For the third term, i.e.
\[\begin{align}
  & 80{{m}^{2}} \\
 & \Rightarrow 2\times 2\times 2\times 2\times 5\times {{m}^{2}} \\
\end{align}\]
For all the three terms, the greatest factors that are common are
\[\begin{align}
  & 2\times 2\times 5 \\
 & \Rightarrow 20 \\
\end{align}\]

Therefore, the greatest common factor of the terms \[140n,140{{m}^{2}},80{{m}^{2}}\] is \[20\].

Note: In this question, always start with the lowest prime factor only. Starting with a factor other than the lowest one may be troublesome. Writing \[\left( 3\times 12 \right)=36\] is factorization and not the prime factorization.