simplify the sum. d^2-7d+12/d^2-d-6 + d^2+5d+6/d^2-2d-15

Answer:The sum is [tex]\frac{2d^2-5d+24}{d^2-3d-10}[/tex]Step-by-step explanation:The given expression is
[tex]\frac{d^2-7d+12}{d^2-d-6}+\frac{d^2+5d+6}{d^2-2d-15}[/tex]
These are all quadratic trinomials, so we split each of the middle terms to obtain,[tex]\frac{d^2-4d-3d+12}{d^2-3d+2d-6}+\frac{d^2+2d+3d+6}{d^2-5d+3d-15}[/tex]
We factor to obtain,
[tex]\frac{d(d-4)-3(d-4)}{d(d-3)+2(d-3)}+\frac{d(d+2)+3(d+2)}{d(d-5)+3(d-5)}[/tex]
We factor further to obtain,[tex]\frac{(d-4)(d-3)}{(d-3)(d+2)}+\frac{(d+2)(d+3)}{(d-5)(d+3)}[/tex]
We cancel out the common factors to get,
[tex]\frac{(d-4)}{(d+2)}+\frac{(d+2)}{(d-5)}[/tex]
The least common denominator is [tex](d+2)(d-5)[/tex].
We collect LCD to get
[tex]\frac{(d-4)(d-5)+(d+2)(d+2)}{(d+2)(d-5)}[/tex]
We now expand to get,
[tex]\frac{d^2-5d-4d+20+d^2+2d+2d+4}{d^2-5d+2d-10}[/tex]
We simplify to get,
[tex]\frac{d^2-9d+20+d^2+4d+4}{d^2-3d-10}[/tex]
We simplify further to get,
[tex]\frac{2d^2-5d+24}{d^2-3d-10}[/tex]