Question

Please help with the question below. Thanks. :)

Answer 1

A. x – 5; where x doesn't equal –2
B. x – 5; where x doesn't equal 5
C. 1/x - 2; where x doesn't equal 2
D. x - 2

Answer 2

The equation is\[\frac{x^2-3x-10}{x+2}\]Try to factor the numerator.\[x^2-3x-10=(x+a)(x+b)\]where (a + b) = -3 and ab = -10.

Answer 3

Hint: I'm guessing that one of the factors will be (x + 2) so it will cancel with the denominator.

Answer 4

First factorise the expression in the numerator.
\[\frac{x^2-3x-10}{x+2} = \frac{(x-5)(x+2)}{x+2} = ...?\]
Then, cancel the common factor, you'll get the simplified form.
After that, put x+2 =0, solve x. The value you get should be rejected as that would make the denominator equal to 0