Question 85478
{{{((n^2 - 9*n + 20)/(2*n^2))*((n^2 + 5*n)/(n^2 - 25))}}} Start with the given expression (note: it should be {{{n^2 - 9*n + 20}}} not {{{n^2 - 9*n - 20}}})



{{{(((n-4)(n-5))/(2*n^2))*((n^2 + 5*n)/(n^2 - 25))}}} Factor the first numerator (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/quadratic-factoring.solver>solver</a>)





{{{(((n-4)(n-5))/(2n^2))*((n(n+5))/(n^2 - 25))}}} Factor the second numerator 



{{{(((n-4)(n-5))/(2n^2))*((n(n+5))/((n+5)(n-5)))}}} Factor the second denominator 


{{{(((n-4)highlight((n-5)))/(2*n^2))*((n*highlight((n+5))/(highlight((n+5))highlight((n-5)))))}}} Notice we have these common terms


{{{(((n-4)cross((n-5)))/(2*n^2))*((n*cross((n+5))/(cross((n+5))cross((n-5)))))}}} They divide and cancel out  


{{{((n-4)/(2n^2))(n/1)}}}

Now divide {{{n}}} and {{{2n^2}}} to get {{{1/2n}}}


{{{(n-4)/(2n)}}}

So the expression


{{{((n^2 - 9*n + 20)/(2*n^2))*((n^2 + 5*n)/(n^2 - 25))}}}


simplifies to


{{{(n-4)/(2n)}}}




As always, we can verify our answer. We can graph the original expression


{{{((n^2 - 9*n + 20)/(2*n^2))*((n^2 + 5*n)/(n^2 - 25))}}}


as a function of x and y like this


{{{ graph( 300, 200, -6, 5, -10, 10, ((x^2 - 9*x + 20)/(2*x^2))*((x^2 + 5*x)/(x^2 - 25))) }}} graph of  {{{y=((x^2 - 9*x + 20)/(2*x^2))*((x^2 + 5*x)/(x^2 - 25))}}}


and graph the simplified result


{{{(n-4)/(2n)}}}


{{{ graph( 300, 200, -6, 5, -10, 10,(x-4)/(2x))}}} graph of {{{y=(x-4)/(2x)}}}


Since they produce the same graphs, this means they are equivalent. Since they are equivalent, this means that our answer has been verified.