The given expression is
[tex]\frac{(x^2-x-6)}{(2x^2+x-6)}\cdot\frac{(2x^2+7x-15)}{(x^2-9)}[/tex]We have to factor all the expressions, one by one.
Let's use the quadratic formula to find the solutions.
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where a = 1, b = -1, and c = -6.
[tex]\begin{gathered} x=\frac{-(-1)\pm\sqrt[]{(-1)^2-4(1)(-6)}}{2(1)}=\frac{1\pm\sqrt[]{1+24}}{2}=\frac{1\pm\sqrt[]{25}_{}}{2}=\frac{1\pm5}{2} \\ x_1=\frac{1+5_{}}{2}=\frac{6}{2}=3\to(x-3) \\ x_2=\frac{1-5}{2}=\frac{-4}{2}=-2\to(x+2) \end{gathered}[/tex]So, the expression in factored form is (x+2)(x-3).
Where a = 2, b = 1, and c = -6. Let's repeat the process.
[tex]\begin{gathered} x=\frac{-1\pm\sqrt[]{1^2-4(2)(-6)}}{2(2)}=\frac{-1\pm\sqrt[]{1+48}}{4}=\frac{-1\pm\sqrt[]{49}}{4}=\frac{-1\pm7}{4} \\ x_1=\frac{-1+7}{4}=\frac{6}{4}=\frac{3}{2}\to(2x-3) \\ x_2=\frac{-1-7_{}}{4}=-\frac{8}{4}=-2\to(x+2) \end{gathered}[/tex]So, the expression in factored form is (2x-3)(x+2).
Where a = 2, b = 7, and c = -15.
[tex]\begin{gathered} x=\frac{-7\pm\sqrt[]{7^2-4(2)(-15)}}{2(2)}=\frac{-7\pm\sqrt[]{49+120}}{4}=\frac{-7\pm\sqrt[]{169}}{4}=\frac{-7\pm13}{4} \\ x_1=\frac{-7+13}{4}=\frac{6}{4}=\frac{3}{2}\to(2x-3) \\ x_2=\frac{-7-13}{4}=\frac{-20}{4}=-5\to(x+5) \end{gathered}[/tex]The expression in factored form is (2x-3)(x+5).
In this case, we have a difference between perfect squares, which can be solved using the following.
[tex]a^2-b^2=(a+b)(a-b)[/tex]Where a^2 = x^2 and b^2 = 9.
[tex]b^2=9\to b=\sqrt[]{9}\to b=3[/tex]So, the expression in factored form is (x+3)(x-3).
Once we have all the factored forms, we simplify.
[tex]\frac{(x+2)(x-3)}{(2x-3)(x+2)}\cdot\frac{(2x-3)(x+5)}{(x+3)(x-3)}=\frac{x+5}{x+3}[/tex]