Question

Find the domain:
I think factoring may be involved in this...
\[g(x)=\frac{ 3 }{ x ^{2}-2x-15 }\]

Answer 1

well you need to factor the denominator and find the values of x that make the denominator zero.

the values of x will be the restrictions in the domain.

the domain is all real x, except the restrictions you find.

hope that helps

Answer 2

Before I can find the restrictions, is it mandatory that I factor the denominator first?

Answer 3

well the restrictions occur in the denominator... so to find them factoring your quadratic is an easy solution...

of if that's a problem you can use the general quadratic formula

Answer 4

the factored form is

\[x^2 - 2x -15 = (x - 5)(x +3)\]

so find the values of x that make the equation equal zero...

they will be the restrictions

Answer 5

If I remember doing this correctly, then this factored would be:

\(\ \dfrac{3}{x^2 -5x + 3x - 15} \)


\(\ \dfrac{3}{x(x - 5) + 3(x - 5)} \)

Then, \(\ \dfrac{3}{(x - 5)(x + 3)}\)

The restrictions are; x = 5, x = -3

\(\ \dfrac{3}{5^2 - 2(5) - 15} \implies \dfrac{3}{25 - 10 - 15}\implies \dfrac{3}{0} \)

Answer 6

oh my goodness you guys ares geniuses e.e thank you...I think I'll try the next problem myself