Question

Please help with this q

Answer 1

$\frac{2 {x}^{2} - 2x }{x - 1} \div \frac{2x + 8}{x - 3} \\ = \frac{2x {}^{2} - 2x}{x - 1} \times \frac{x - 3}{2x + 8} \\ = \frac{(2 {x}^{2} - 2x)(x - 3) }{(x - 1)(2x + 8)}$

So:

1) x - 1 ≠ 0

x ≠ 1

2) 2x + 8 ≠ 0

2x ≠ -8

x ≠ -4

(B)