9.
Option (C) is the correct one.
[tex] \frac{ {x}^{2} - 5x + 6 }{ {x}^{2} + 6x + 9} \times \frac{ {x}^{2} + 7x + 12 }{ {x}^{2} - 9} \\ \\ = \frac{(x - 2)(x - 3)}{ {(x + 3)}^{2} } \times \frac{(x + 4)(x + 3)}{(x + 3)(x - 3)} \\ = \frac{(x - 2)(x + 4)}{ {(x + 3)}^{2} } [/tex]
10. I'll go for Option (B).
Reason:
[tex] \frac{2 {x}^{2} + 16x + 24 }{ {x}^{2} + 2x + 4 } \div \frac{ {x}^{2} + 7x + 6} { {x}^{2} - 4 } \\ \\ = \frac{ 2{x}^{2} + 16x + 24 }{ {x}^{2} + 2x + 4 } \times \frac{ {x}^{2} - 4}{ {x}^{2} + 7x + 6} \\ \\ = \frac{2(x + 2)(x + 6)}{ {x}^{2} + 2x + 4} \times \frac{(x - 2)(x + 2)}{(x - 1)(x + 6)} \\ \\ [/tex]
I think we have a problem here in the denominator of the first multiple.
The term
[tex] {x}^{2} + 2x + 4[/tex]
cannot be broken down into into real terms.
You can check it yourself,
Using,
[tex] {b}^{2} - 4ac[/tex]
Where b is the coefficient of x and c is the constant term in a given quadratic equation. If the value is smaller than 0 then there are no real roots of the equation.
Since we aren't don't provide any such options I'm gonna change b= 4 here just for the sake of getting answer.
[tex] \frac{2(x + 2)(x + 6)}{ {x}^{2} + 4x + 4} \times \frac{(x - 2)(x + 2)}{(x - 1)(x - 6)} \\ \\ = \frac{2(x + 2)(x + 6)}{(x + 2)(x + 2)} \times \frac{(x - 2)(x + 2)}{(x - 1)(x + 6)} \\ \\ = \frac{2(x - 2)}{(x - 1)} [/tex]
