Answer:
Option 2 is correct that is [tex]\frac{x+2}{x+9}[/tex] and [tex]x\neq-9[/tex]and [tex]x\neq-12[/tex]
Step-by-step explanation:
We have been given with the expression [tex]\frac{x^2-10x-24}{x^2-3x-108}[/tex]
We will simplify the given expression by factorisation we will get [tex]\frac{x^2-12x+2x-24}{x^2-12x+9x-108}[/tex]
Taking x as common from first two terms in numerator and 2 from last two terms in numerator
Similarly, take x common from first two terms and 9 from last two terms in denominator we will get
[tex]\frac{x(x-12)+2(x-12)}{x(x-12)+9(x-12)}[/tex]
After arranging the terms we will get [tex]\frac{(x+2)(x-12)}{(x+9)(x-12)}[/tex]
Taking out the common factor which is (x-12) from numerator and denominator it will get cancelled we will get
[tex]\frac{(x+2)}{(x+9)}[/tex]
Hence, Option 2 is correct that is [tex]\frac{x+2}{x+9}[/tex] and [tex]x\neq-9[/tex]and [tex]x\neq-12[/tex]
Answer:
x + 2 over x + 9
x doesn’t equal to -9
x doesn’t equal to 12
Step-by-step explanation:
