Algebra Examples

$\frac{y^{2} - 36}{8 y^{2} + 96 y + 288}$

Step 1

Set the denominator in $\frac{y^{2} - 36}{8 y^{2} + 96 y + 288}$ equal to $0$ to find where the expression is undefined.

$8 y^{2} + 96 y + 288 = 0$

Step 2

Solve for $y$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Factor $8$ out of $8 y^{2} + 96 y + 288$ .

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Step 2.1.1.1

Factor $8$ out of $8 y^{2}$ .

$8 (y^{2}) + 96 y + 288 = 0$

Step 2.1.1.2

Factor $8$ out of $96 y$ .

$8 (y^{2}) + 8 (12 y) + 288 = 0$

Step 2.1.1.3

Factor $8$ out of $288$ .

$8 y^{2} + 8 (12 y) + 8 \cdot 36 = 0$

Step 2.1.1.4

Factor $8$ out of $8 y^{2} + 8 (12 y)$ .

$8 (y^{2} + 12 y) + 8 \cdot 36 = 0$

Step 2.1.1.5

Factor $8$ out of $8 (y^{2} + 12 y) + 8 \cdot 36$ .

$8 (y^{2} + 12 y + 36) = 0$

Step 2.1.2

Factor using the perfect square rule.

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Step 2.1.2.1

Rewrite $36$ as $6^{2}$ .

$8 (y^{2} + 12 y + 6^{2}) = 0$

Step 2.1.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 y = 2 \cdot y \cdot 6$

Step 2.1.2.3

Rewrite the polynomial.

$8 (y^{2} + 2 \cdot y \cdot 6 + 6^{2}) = 0$

Step 2.1.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = y$ and $b = 6$ .

$8 {(y + 6)}^{2} = 0$

Step 2.2

Divide each term in $8 {(y + 6)}^{2} = 0$ by $8$ and simplify.

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Step 2.2.1

Divide each term in $8 {(y + 6)}^{2} = 0$ by $8$ .

$\frac{8 {(y + 6)}^{2}}{8} = \frac{0}{8}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $8$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{8 {(y + 6)}^{2}}{8} = \frac{0}{8}$

Step 2.2.2.1.2

Divide ${(y + 6)}^{2}$ by $1$ .

${(y + 6)}^{2} = \frac{0}{8}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $0$ by $8$ .

${(y + 6)}^{2} = 0$

Step 2.3

Set the $y + 6$ equal to $0$ .

$y + 6 = 0$

Step 2.4

Subtract $6$ from both sides of the equation.

$y = - 6$

Step 3