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# rational expressions g^2-6g-55 / gk^3 +k / k^2 + k-42

need to find the domain for both expressions

g^2 - 6g-55 divided by g

k^3 + k divided by K^2 + k - 42

(g^2 - 6g - 55)/g

Can’t divide by zero, so domain is g ≠ 0.

In Interval Notation the domain is (-∞,0) ∪ (0,∞).

(k^3 + k)/(k^2 + k - 42) =
(k(k^2 + 1)) / [(k+7)(k-6)]

Check factoring: (k+7)(k-6) = k*k - 6k + 7k - 42 √

Can’t divide by zero, so domain is k ≠ -7 AND k ≠ 6.

In Interval Notation the domain is (-∞,-7) ∪ (-7,6) ∪ (6,∞).
Hi Margaret;
For the first equation, the denominator, also known as the divisor, is g.  No denominator can be zero.  Henceforth, this is the only limitation on the domain...
(-infinity,0)(0,+infinity)
Please note that the parentheses, as opposed to brackets, indicate that the zero is NOT included.

For the second equation, let's look at the denominator/divisor...
k2 + k - 42
Let's factor.
For the FOIL...
FIRST must be...(k)(k)=k2
OUTER and INNER must add-up to 1k or k.
LAST...(6)(-7).
(k+6)(k-7)
Let's FOIL...
FIRST...(k)(k)=k2
OUTER...(k)(-7)=-7k
INNER...(6)(k)=6k
LAST...(6)(-7)=-42
k2-7k+6k-42
k2-7k-42
(k+6)(k-7)
Neither parenthetical equation can result in zero...
k+6=0
k=-6
and
k-7=0
k=7
Domain is...
(-infinity,-6)(-6,7)(7,+infinity)
Parentheses indicate that neither -6 nor 7 are included.
AS PER STEVE'S COMMENT, I MADE THE CORRECTION HEREIN.

Bracket convention is:
1. "open bracket", i.e., parenthesis, means NOT included and
2. "closed bracket", i.e., [ or ], means IS included.

I think you have it backwards.

One way to remember is to write (-∞,∞) for all real numbers. Since ±∞ are not numbers they can't be included; hence the parentheses to indicate an open interval.
My apologies.  I have it backwards.