Precalculus Examples

$(x + 2) (x - 2) = 15 + (x + 1) (x - 7)$

Step 1

Simplify $(x + 2) (x - 2)$ .

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Step 1.1

Rewrite.

$0 + 0 + (x + 2) (x - 2) = 15 + (x + 1) (x - 7)$

Step 1.2

Simplify by adding zeros.

$(x + 2) (x - 2) = 15 + (x + 1) (x - 7)$

Step 1.3

Expand $(x + 2) (x - 2)$ using the FOIL Method.

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Step 1.3.1

Apply the distributive property.

$x (x - 2) + 2 (x - 2) = 15 + (x + 1) (x - 7)$

Step 1.3.2

Apply the distributive property.

$x \cdot x + x \cdot - 2 + 2 (x - 2) = 15 + (x + 1) (x - 7)$

Step 1.3.3

Apply the distributive property.

$x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2 = 15 + (x + 1) (x - 7)$

Step 1.4

Simplify terms.

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Step 1.4.1

Combine the opposite terms in $x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2$ .

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Step 1.4.1.1

Reorder the factors in the terms $x \cdot - 2$ and $2 x$ .

$x \cdot x - 2 x + 2 x + 2 \cdot - 2 = 15 + (x + 1) (x - 7)$

Step 1.4.1.2

Add $- 2 x$ and $2 x$ .

$x \cdot x + 0 + 2 \cdot - 2 = 15 + (x + 1) (x - 7)$

Step 1.4.1.3

Add $x \cdot x$ and $0$ .

$x \cdot x + 2 \cdot - 2 = 15 + (x + 1) (x - 7)$

Step 1.4.2

Simplify each term.

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Step 1.4.2.1

Multiply $x$ by $x$ .

$x^{2} + 2 \cdot - 2 = 15 + (x + 1) (x - 7)$

Step 1.4.2.2

Multiply $2$ by $- 2$ .

$x^{2} - 4 = 15 + (x + 1) (x - 7)$

Step 2

Simplify $15 + (x + 1) (x - 7)$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Expand $(x + 1) (x - 7)$ using the FOIL Method.

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Step 2.1.1.1

Apply the distributive property.

$x^{2} - 4 = 15 + x (x - 7) + 1 (x - 7)$

Step 2.1.1.2

Apply the distributive property.

$x^{2} - 4 = 15 + x \cdot x + x \cdot - 7 + 1 (x - 7)$

Step 2.1.1.3

Apply the distributive property.

$x^{2} - 4 = 15 + x \cdot x + x \cdot - 7 + 1 x + 1 \cdot - 7$

Step 2.1.2

Simplify and combine like terms.

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Step 2.1.2.1

Simplify each term.

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Step 2.1.2.1.1

Multiply $x$ by $x$ .

$x^{2} - 4 = 15 + x^{2} + x \cdot - 7 + 1 x + 1 \cdot - 7$

Step 2.1.2.1.2

Move $- 7$ to the left of $x$ .

$x^{2} - 4 = 15 + x^{2} - 7 \cdot x + 1 x + 1 \cdot - 7$

Step 2.1.2.1.3

Multiply $x$ by $1$ .

$x^{2} - 4 = 15 + x^{2} - 7 x + x + 1 \cdot - 7$

Step 2.1.2.1.4

Multiply $- 7$ by $1$ .

$x^{2} - 4 = 15 + x^{2} - 7 x + x - 7$

Step 2.1.2.2

Add $- 7 x$ and $x$ .

$x^{2} - 4 = 15 + x^{2} - 6 x - 7$

Step 2.2

Subtract $7$ from $15$ .

$x^{2} - 4 = x^{2} - 6 x + 8$

Step 3

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} - 6 x + 8 = x^{2} - 4$

Step 4

Move all terms containing $x$ to the left side of the equation.

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Step 4.1

Subtract $x^{2}$ from both sides of the equation.

$x^{2} - 6 x + 8 - x^{2} = - 4$

Step 4.2

Combine the opposite terms in $x^{2} - 6 x + 8 - x^{2}$ .

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Step 4.2.1

Subtract $x^{2}$ from $x^{2}$ .

$- 6 x + 8 + 0 = - 4$

Step 4.2.2

Add $- 6 x + 8$ and $0$ .

$- 6 x + 8 = - 4$

Step 5

Move all terms not containing $x$ to the right side of the equation.

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Step 5.1

Subtract $8$ from both sides of the equation.

$- 6 x = - 4 - 8$

Step 5.2

Subtract $8$ from $- 4$ .

$- 6 x = - 12$

Step 6

Divide each term in $- 6 x = - 12$ by $- 6$ and simplify.

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Step 6.1

Divide each term in $- 6 x = - 12$ by $- 6$ .

$\frac{- 6 x}{- 6} = \frac{- 12}{- 6}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $- 6$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{- 6 x}{- 6} = \frac{- 12}{- 6}$

Step 6.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 12}{- 6}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Divide $- 12$ by $- 6$ .

$x = 2$