Pre-Algebra Examples

$\frac{z^{2}}{6} - \frac{3 z}{2} - 6 = 0$

Step 1

Multiply each term in $\frac{z^{2}}{6} - \frac{3 z}{2} - 6 = 0$ by $6$ to eliminate the fractions.

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Step 1.1

Multiply each term in $\frac{z^{2}}{6} - \frac{3 z}{2} - 6 = 0$ by $6$ .

$\frac{z^{2}}{6} \cdot 6 - \frac{3 z}{2} \cdot 6 - 6 \cdot 6 = 0 \cdot 6$

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify each term.

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Step 1.2.1.1

Cancel the common factor of $6$ .

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Step 1.2.1.1.1

Cancel the common factor.

$\frac{z^{2}}{6} \cdot 6 - \frac{3 z}{2} \cdot 6 - 6 \cdot 6 = 0 \cdot 6$

Step 1.2.1.1.2

Rewrite the expression.

$z^{2} - \frac{3 z}{2} \cdot 6 - 6 \cdot 6 = 0 \cdot 6$

Step 1.2.1.2

Cancel the common factor of $2$ .

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Step 1.2.1.2.1

Move the leading negative in $- \frac{3 z}{2}$ into the numerator.

$z^{2} + \frac{- 3 z}{2} \cdot 6 - 6 \cdot 6 = 0 \cdot 6$

Step 1.2.1.2.2

Factor $2$ out of $6$ .

$z^{2} + \frac{- 3 z}{2} \cdot (2 (3)) - 6 \cdot 6 = 0 \cdot 6$

Step 1.2.1.2.3

Cancel the common factor.

$z^{2} + \frac{- 3 z}{2} \cdot (2 \cdot 3) - 6 \cdot 6 = 0 \cdot 6$

Step 1.2.1.2.4

Rewrite the expression.

$z^{2} - 3 z \cdot 3 - 6 \cdot 6 = 0 \cdot 6$

Step 1.2.1.3

Multiply $3$ by $- 3$ .

$z^{2} - 9 z - 6 \cdot 6 = 0 \cdot 6$

Step 1.2.1.4

Multiply $- 6$ by $6$ .

$z^{2} - 9 z - 36 = 0 \cdot 6$

Step 1.3

Simplify the right side.

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Step 1.3.1

Multiply $0$ by $6$ .

$z^{2} - 9 z - 36 = 0$

Step 2

Factor $z^{2} - 9 z - 36$ using the AC method.

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Step 2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 36$ and whose sum is $- 9$ .

$- 12, 3$

Step 2.2

Write the factored form using these integers.

$(z - 12) (z + 3) = 0$

Step 3

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

$z - 12 = 0$

$z + 3 = 0$

Step 4

Set $z - 12$ equal to $0$ and solve for $z$ .

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Step 4.1

Set $z - 12$ equal to $0$ .

$z - 12 = 0$

Step 4.2

Add $12$ to both sides of the equation.

$z = 12$

Step 5

Set $z + 3$ equal to $0$ and solve for $z$ .

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Step 5.1

Set $z + 3$ equal to $0$ .

$z + 3 = 0$

Step 5.2

Subtract $3$ from both sides of the equation.

$z = - 3$

Step 6

The final solution is all the values that make $(z - 12) (z + 3) = 0$ true.

$z = 12, - 3$