```Question 169800

{{{((x^2+2x-15)/(x^2-9))((x^2+4x+3)/(x^2+6x+5))}}} Multiply the first fraction {{{(x^2+2x-15)/(x^2-9)}}} by the reciprocal of the second fraction {{{(x^2+6x+5)/(x^2+4x+3)}}}.

{{{(((x+5)(x-3))/(x^2-9))((x^2+4x+3)/(x^2+6x+5))}}} Factor {{{x^2+2x-15}}} to get {{{(x+5)(x-3)}}}.

{{{(((x+5)(x-3))/((x-3)(x+3)))((x^2+4x+3)/(x^2+6x+5))}}} Factor {{{x^2-9}}} to get {{{(x-3)(x+3)}}}.

{{{(((x+5)(x-3))/((x-3)(x+3)))(((x+3)(x+1))/(x^2+6x+5))}}} Factor {{{x^2+4x+3}}} to get {{{(x+3)(x+1)}}}.

{{{(((x+5)(x-3))/((x-3)(x+3)))(((x+3)(x+1))/((x+5)(x+1)))}}} Factor {{{x^2+6x+5}}} to get {{{(x+5)(x+1)}}}.

{{{((x+5)(x-3)(x+3)(x+1))/((x-3)(x+3)(x+5)(x+1))}}} Combine the fractions.

{{{(highlight((x+5))highlight((x-3))highlight((x+3))highlight((x+1)))/(highlight((x-3))highlight((x+3))highlight((x+5))highlight((x+1)))}}} Highlight the common terms.

{{{(cross((x+5))cross((x-3))cross((x+3))cross((x+1)))/(cross((x-3))cross((x+3))cross((x+5))cross((x+1)))}}} Cancel out the common terms.

{{{1}}} Simplify.

So {{{((x^2+2x-15)/(x^2-9))/((x^2+6x+5)/(x^2+4x+3))}}} simplifies to {{{1}}}.

In other words, {{{((x^2+2x-15)/(x^2-9))/((x^2+6x+5)/(x^2+4x+3))=1}}} where {{{x<>-5}}}, {{{x<>-3}}}, {{{x<>-1}}}, or {{{x<>3}}}
```