Question

(x^2-5x-14)/(x^2-9) multiplied (x^2-2x-3)/(x^2-6x-7)

Answer 1

\[\frac{x^2-5x-14}{x^2-9}\times\frac{x^2-2x-3}{x^2-6x-7}\]
\[\frac{(x-7)(x+2)}{(x-3)(x+3)}\times\frac{(x-3)(x+1)}{(x-7)(x+1)}\]
\[\frac{(x+2)}{(x+3)}\]

Answer 2

\[(x^2-5x-14)/(x^2-9) *(x^2-2x-3)/(x^2-6x-7)\]\[(x^2-5x-14)*(x^2-2x-3)/(x^2-9)*(x^2-6x-7)\]Now, let's factor these polynomial to make our life easier.
Let be \[P(x)=(x^2-5x-14)\]\[Δ=(-5)^2-4(-14)=81\]\[x_1,2=(-(-5)\pm \sqrt{81})/2*1=(5\pm9)/2\]So P(x)=\[(x-7)(x+2)Q(x)=x^2-2x-3\[Δ=(-2)^2-4(-3)=16\]\[x_1,2=(-(-2)\pm \sqrt{16})/2*1\]Eventually Q(x)=(x-3)(x+1)\]
\[x^2-9\] is an identity so it is equal to\[(x-3)(x+3)\]Finally \[f(x)=x^2-6x-7\]\[Δ=(-6)^2-4(-7)=64\]\[x_1,2=(-(-6)\pm \sqrt{64})/2*1\]\[f(x)=(x-7)(x+1)\]
\[(x-7)(x+2)*(x-3)(x+1)\over(x-3)(x+3)*(x-7)(x+1)\]\[(x+2)\over(x+3)\]