need to find the domain for both expressions

g^2 - 6g-55 divided by g

k^3 + k divided by 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

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Westford, MA

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

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

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.

(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,∞).

Middletown, CT

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...

k^{2} + k - 42

Let's factor.

For the FOIL...

FIRST must be...(k)(k)=k^{2}

OUTER and INNER must add-up to 1k or k.

LAST...(6)(-7).

(k+6)(k-7)

Let's FOIL...

FIRST...(k)(k)=k^{2}

OUTER...(k)(-7)=-7k

INNER...(6)(k)=6k

LAST...(6)(-7)=-42

k^{2}-7k+6k-42

k^{2}-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.

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