Answer
$\dfrac{4x^2+1}{3x^2-1}$
Work Step by Step
Multiply the numerator and the denominator by $x^2$.
$=\dfrac{x^2(4+\frac{1}{x^2})}{x^2(3-\frac{1}{x^2})}$
Use distributive property.
$=\dfrac{4x^2+x^2\cdot \frac{1}{x^2}}{3x^2-x^2\cdot \frac{1}{x^2}}$
Simplify.
$=\dfrac{4x^2+1}{3x^2-1}$
Hence, the lowest term is:
$\dfrac{4x^2+1}{3x^2-1}$
