Answer
$-\dfrac{(x+3)(3x-1)}{(x^2+1)^2}$.
Work Step by Step
Use distributive property.
$\dfrac{(x^2+1)\cdot 3-(3x+4)\cdot 2x}{(x^2+1)^2}\\
=\dfrac{3x^2+ 3-6x^2-8x}{(x^2+1)^2}$
Simplify by combining like terms.
$=\dfrac{-3x^2-8x+3}{(x^2+1)^2}$
Factor the numerator.
Factor out $-1$:
$=\dfrac{-(3x^2+8x-3)}{(x^2+1)^2}$
$=-\dfrac{3x^2+8x-3}{(x^2+1)^2}$
With $a=3$ and $c=-3$, $ac=3(-3)=-9$.
Look for factors of $-9$ whose sum is equal to $b=8$.
The factors are $9$ and $-1$.
Rewrite the middle term $8x$ as $9x-1x$ to obtain:
$=-\dfrac{3x^2+9x-1x-3}{(x^2+1)^2}$
$=-\dfrac{(3x^2+9x)+(-1x-3)}{(x^2+1)^2}$
Factor out the GCF (Greatest Common Factor) in each group to obtain:
$=-\dfrac{3x(x+3)-1(x+3)}{(x^2+1)^2}$
$=-\dfrac{(x+3)(3x-1)}{(x^2+1)^2}$
The numerator and denominator have no common factors.
Hence, the lowest term is $-\dfrac{(x+3)(3x-1)}{(x^2+1)^2}$.
