Question 78008
Does your problem look like this?

{{{(x^2+5x-14)/(x^2+10x+21)}}}

If this is the case, I'm afraid you don't have the answer (despite what the other tutor said. He must have quickly glanced at it).

First factor the numerator

*[invoke factoring_quadratics 1, 5, -14]





Factor the denominator



*[invoke factoring_quadratics 1, 10, 21]

So when we factor the numerator and denominator we get:

{{{((x-2)(x+7))/((x+3)(x+7))}}}



{{{((x-2)cross((x+7)))/((x+3)cross((x+7)))}}} Cancel like terms


{{{(x-2)/(x+3)}}} So this is your answer. 

In other words, {{{(x^2+5x-14)/(x^2+10x+21)}}} reduces to {{{(x-2)/(x+3)}}}


note: you were close though. You just had an extra x in your denominator.


Notice if you graph {{{(x^2+5x-14)/(x^2+10x+21)}}} you get

{{{ graph( 300, 200, -6, 5, -10, 10, (x^2+5x-14)/(x^2+10x+21)) }}} graph of {{{(x^2+5x-14)/(x^2+10x+21)}}}

it should be the same as the graph of {{{(x-2)/(x+3)}}}

{{{ graph( 300, 200, -6, 5, -10, 10, (x-2)/(x+3)) }}} graph of {{{(x-2)/(x+3)}}}

We can see that they are equal. So this verifies our answer.