Answer:
[tex]\frac{2(x+3)}{x+7}[/tex]
Step-by-step explanation:
Given
[tex]\frac{2x^2-18}{x^2+4x-21}[/tex] ← factorise numerator and denominator
2x² - 18 ( factor out 2 from each term )
= 2(x² - 9) ← difference of 2 squares
= 2(x - 3)(x + 3)
and
x² + 4x - 21 = (x + 7)(x - 3) , then
= [tex]\frac{2(x-3)(x+3)}{(x+7)(x-3)}[/tex] ← cancel common factor (x - 3) on numerator/ denominator
= [tex]\frac{2(x+3)}{x+7}[/tex]
