Question 85471
{{{((2*x^2 -x - 3)/(3*x^2 + 7*x + 4))*((3*x^2 - 11*x - 20)/(4*x^2 - 9))}}} Start with the given expression



{{{(((x+1)(2*x-3))/(3*x^2 + 7*x + 4))*((3*x^2 - 11*x - 20)/(4*x^2 - 9))}}} 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>)



{{{(((x+1)(2*x-3))/((x+1)(3*x+4)))*((3*x^2 - 11*x - 20)/(4*x^2 - 9))}}} Factor the first denominator 



{{{(((x+1)(2*x-3))/((x+1)(3*x+4)))*(((3*x+4)(x-5))/(4*x^2 - 9))}}} Factor the second numerator 



{{{(((x+1)(2*x-3))/((x+1)(3*x+4)))*(((3*x+4)(x-5))/((2*x+3)(2*x-3)))}}} Factor the second denominator 



{{{((highlight((x+1))highlight((2*x-3)))/(highlight((x+1))highlight((3*x+4))))*((highlight((3*x+4))(x-5))/((2*x+3)highlight((2*x-3))))}}} Notice we have these common terms



{{{((cross((x+1))cross((2*x-3)))/(cross((x+1))cross((3*x+4))))*((cross((3*x+4))(x-5))/((2*x+3)cross((2*x-3))))}}} They divide and cancel out



Leaving you with this:

{{{((x-5)/(2*x+3))}}}



So the expression


{{{((2*x^2 -x - 3)/(3*x^2 + 7*x + 4))*((3*x^2 - 11*x - 20)/(4*x^2 - 9))}}}


simplifies to


{{{((x-5)/(2*x+3))}}}



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


{{{((2*x^2 -x - 3)/(3*x^2 + 7*x + 4))*((3*x^2 - 11*x - 20)/(4*x^2 - 9))}}}


as a function of y like this


{{{ graph( 300, 200, -6, 5, -10, 10, ((2*x^2 -x - 3)/(3*x^2 + 7*x + 4))*((3*x^2 - 11*x - 20)/(4*x^2 - 9))) }}} graph of  {{{y=((2*x^2 -x - 3)/(3*x^2 + 7*x + 4))*((3*x^2 - 11*x - 20)/(4*x^2 - 9))}}}


and graph the simplified result


{{{((x-5)/(2*x+3))}}} 


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


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