Question 78009
{{{(2x^2+7x+3)/(4x^2-1)}}}

Factor the numerator:


*[invoke quadratic_factoring 2, 7, 3]


Now factor the denominator:


*[invoke quadratic_factoring 4, 0, -1]


So after factoring we get

{{{((2x+1)(x+3))/((2x+1)(2x-1))}}}

{{{(cross((2x+1))(x+3))/(cross((2x+1))(2x-1))}}} Cancel like terms


So {{{(2x^2+7x+3)/(4x^2-1)}}} reduces to {{{(x+3)/(2x-1)}}}

note: you were close, but had the wrong sign. You can always check your answers when you factor with this <a href="http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/quadratic-factoring.solver"> solver </a>. Also, you can check your simplification by graphing.