Question

I need help please? thank you :)

Answer 1

${\qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

$\qquad \sf \dashrightarrow \: \cfrac{ {z}^{2} - 14z + 48 }{ {z}^{2} + 6x - 27} ÷ \cfrac{z -8}{z-3}$

$\qquad \sf \dashrightarrow \: \cfrac{ {z}^{2} - 8z - 6z+ 48 }{ {z}^{2} + 9z- 3z- 27} ÷ \cfrac{z -8}{z-3}$

$\qquad \sf \dashrightarrow \: \cfrac{ {z(}^{} z- 8) - 6(z - 8) }{ {z}^{} (z+ 9)- 3(z + 9)} ÷ \cfrac{z -8}{z-3}$

$\qquad \sf \dashrightarrow \: \cfrac{ {(}^{} z- 8) (z - 6) }{ {}^{} (z+ 9)(z - 3)} ÷ \cfrac{z -8}{z-3}$

$\qquad \sf \dashrightarrow \: \cfrac{ (z - 6) }{ {}^{} (z+ 9)}$

Answer 2

Break into two parts and simplify

$\\ \rm\dashrightarrow \dfrac{z^2-14z+48}{z^2+6z-27}$

$\\ \rm\dashrightarrow \dfrac{z^2-8z-6z+48}{z^2+9z-3z-27}$

$\\ \rm\dashrightarrow \dfrac{(z-8)(z-6)}{(z+9)(z-3)}$

Now divide by the denominator

$\\ \rm\dashrightarrow \dfrac{(z-8)(z-6)(z-3)}{(z+9)(z-3)(z-8)}$

$\\ \rm\dashrightarrow \dfrac{z-6}{z+9}$