Without actually performing the long division,find if $\frac{987}{10500}$ will have a terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.

(A) First,simplify the fraction $\frac{987}{10500}$ by dividing both numerator and denominator by their greatest common divisor,which is $21$.
$\frac{987 \div 21}{10500 \div 21} = \frac{47}{500}$.
Now,find the prime factorization of the denominator $500$:
$500 = 5 \times 100 = 5 \times 10^2 = 5 \times (2 \times 5)^2 = 5 \times 2^2 \times 5^2 = 2^2 \times 5^3$.
According to the theorem,a rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^m \times 5^n$,where $m$ and $n$ are non-negative integers.
Since the denominator $500 = 2^2 \times 5^3$ is in the form $2^m \times 5^n$ (where $m=2$ and $n=3$),the decimal expansion of $\frac{987}{10500}$ is terminating.
To find the decimal value: $\frac{47}{500} = \frac{47 \times 2}{500 \times 2} = \frac{94}{1000} = 0.094$.