Simplifying Algebraic Fraction with Common Powers of 5
The expression given is:
\( \frac{10 \times 5^{n+4} + 125 \times 5^{n+2}}{3 \times 5^{n+3} - 20 \times 5^{n+1}} \)
We can simplify this expression by factoring common terms and any powers of 5 that appear in all terms, with the goal of simplifying the fraction by cancellation where possible.
Let's factor out the greatest power of 5 from the numerator and denominator.
The numerator has \( 5^{n+4} \) and \( 5^{n+2} \), so we can factor out \( 5^{n+2} \):
\( 5^{n+2} \times (10 \times 5^2 + 125) \)
\( = 5^{n+2} \times (10 \times 25 + 125) \)
\( = 5^{n+2} \times (250 + 125) \)
\( = 5^{n+2} \times 375 \)
The denominator has \( 5^{n+3} \) and \( 5^{n+1} \), so we can factor out \( 5^{n+1} \):
\( 5^{n+1} \times (3 \times 5^2 - 20) \)
\( = 5^{n+1} \times (3 \times 25 - 20) \)
\( = 5^{n+1} \times (75 - 20) \)
\( = 5^{n+1} \times 55 \)
After factoring out the common powers of 5, the original expression now looks like this:
\( \frac{5^{n+2} \times 375}{5^{n+1} \times 55} \)
We can cancel out a \( 5^{n+1} \) from the numerator and denominator, leaving us with:
\( \frac{5 \times 375}{55} \)
Divide 375 by 55:
\( \frac{5 \times 375}{55} = \frac{5 \times 15}{11} \)
So, the simplified form of the original expression is:
\( \frac{75}{11} \)
This fraction cannot be simplified further as 75 and 11 are relatively prime (their greatest common divisor is 1).
