Find the Excluded Values (8q+8)/(3q^2-q-14)

$\frac{8 q + 8}{3 q^{2} - q - 14}$

Set the denominator in $\frac{8 q + 8}{3 q^{2} - q - 14}$ equal to $0$ to find where the expression is undefined.

$3 q^{2} - q - 14 = 0$

Solve for $q$ .

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 14 = - 42$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- q$ .

$3 q^{2} - (q) - 14 = 0$

Rewrite $- 1$ as $6$ plus $- 7$

$3 q^{2} + (6 - 7) q - 14 = 0$

Apply the distributive property.

$3 q^{2} + 6 q - 7 q - 14 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 q^{2} + 6 q) - 7 q - 14 = 0$

Factor out the greatest common factor (GCF) from each group.

$3 q (q + 2) - 7 (q + 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $q + 2$ .

$(q + 2) (3 q - 7) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$q + 2 = 0$

$3 q - 7 = 0$

Set $q + 2$ equal to $0$ and solve for $q$ .

Set $q + 2$ equal to $0$ .

$q + 2 = 0$

Subtract $2$ from both sides of the equation.

$q = - 2$

Set $3 q - 7$ equal to $0$ and solve for $q$ .

Set $3 q - 7$ equal to $0$ .

$3 q - 7 = 0$

Solve $3 q - 7 = 0$ for $q$ .

Add $7$ to both sides of the equation.

$3 q = 7$

Divide each term in $3 q = 7$ by $3$ and simplify.

Divide each term in $3 q = 7$ by $3$ .

$\frac{3 q}{3} = \frac{7}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 q}{3} = \frac{7}{3}$

Divide $q$ by $1$ .

$q = \frac{7}{3}$

The final solution is all the values that make $(q + 2) (3 q - 7) = 0$ true.

$q = - 2, \frac{7}{3}$

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$q = - 2, q = \frac{7}{3}$

Find the Excluded Values (8q+8)/(3q^2-q-14)

Related Questions

Support LCMGCF.com by disabling your adblocker.