Find the Excluded Values (2y^2+2)/(y^3-5y^2+y-5)

$\frac{2 y^{2} + 2}{y^{3} - 5 y^{2} + y - 5}$

Set the denominator in $\frac{2 y^{2} + 2}{y^{3} - 5 y^{2} + y - 5}$ equal to $0$ to find where the expression is undefined.

$y^{3} - 5 y^{2} + y - 5 = 0$

Solve for $y$ .

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(y^{3} - 5 y^{2}) + y - 5 = 0$

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

$y^{2} (y - 5) + 1 (y - 5) = 0$

Factor the polynomial by factoring out the greatest common factor, $y - 5$ .

$(y - 5) (y^{2} + 1) = 0$

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

$y - 5 = 0$

$y^{2} + 1 = 0$

Set $y - 5$ equal to $0$ and solve for $y$ .

Set $y - 5$ equal to $0$ .

$y - 5 = 0$

Add $5$ to both sides of the equation.

$y = 5$

Set $y^{2} + 1$ equal to $0$ and solve for $y$ .

Set $y^{2} + 1$ equal to $0$ .

$y^{2} + 1 = 0$

Solve $y^{2} + 1 = 0$ for $y$ .

Subtract $1$ from both sides of the equation.

$y^{2} = - 1$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{- 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \pm i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = i$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - i$

The complete solution is the result of both the positive and negative portions of the solution.

$y = i, - i$

The final solution is all the values that make $(y - 5) (y^{2} + 1) = 0$ true.

$y = 5, i, - i$

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 $0$ .

$y = 5$

Find the Excluded Values (2y^2+2)/(y^3-5y^2+y-5)

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