\(\frac{48-3x^2}{x^2+x-20}\)
remember difference of 2 perfect squares,
a^2-b^2=(a+b)(a-b)
factor them to find ones
48-3x^2=3(16-x^2)=(3)(4+x)(4-x)
x^2+1x-20
find what 2 numbers multiply to -20 and add to 1
-4 and 5
(x-4)(x+5)
now we have
\(\frac{(3)(4-x)(4+x)}{(x-4)(x+5)}\)
we can clear that (x-4) by force undistributing a negative 1 from that (4-x)
(4-x)=(-1)(x-4)
\(\frac{(3)(-1)(x-4)(4+x)}{(x-4)(x+5)}\)=
\(\frac{(3)(-1)(4+x)}{(x+5)}\)=
\(\frac{(-3)(4+x)}{(x+5)}\)=
\(\frac{(-12-3x)}{(x+5)}\)
