Precalculus Examples

$4 (x + h) = 4 (x + h^{2} + 2 + h)$

Step 1

Simplify $4 (x + h)$ .

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Step 1.1

Rewrite.

$0 + 0 + 4 (x + h) = 4 (x + h^{2} + 2 + h)$

Step 1.2

Simplify by adding zeros.

$4 (x + h) = 4 (x + h^{2} + 2 + h)$

Step 1.3

Apply the distributive property.

$4 x + 4 h = 4 (x + h^{2} + 2 + h)$

Step 2

Simplify $4 (x + h^{2} + 2 + h)$ .

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Step 2.1

Apply the distributive property.

$4 x + 4 h = 4 x + 4 h^{2} + 4 \cdot 2 + 4 h$

Step 2.2

Multiply $4$ by $2$ .

$4 x + 4 h = 4 x + 4 h^{2} + 8 + 4 h$

Step 3

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

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Step 3.1

Subtract $4 x$ from both sides of the equation.

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

Step 3.2

Combine the opposite terms in $4 x + 4 h - 4 x$ .

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Step 3.2.1

Subtract $4 x$ from $4 x$ .

$4 h + 0 = 4 h^{2} + 8 + 4 h$

Step 3.2.2

Add $4 h$ and $0$ .

$4 h = 4 h^{2} + 8 + 4 h$

Step 4

Divide each term in $4 h = 4 h^{2} + 8 + 4 h$ by $4$ and simplify.

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Step 4.1

Divide each term in $4 h = 4 h^{2} + 8 + 4 h$ by $4$ .

$\frac{4 h}{4} = \frac{4 h^{2}}{4} + \frac{8}{4} + \frac{4 h}{4}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $4$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{4 h}{4} = \frac{4 h^{2}}{4} + \frac{8}{4} + \frac{4 h}{4}$

Step 4.2.1.2

Divide $h$ by $1$ .

$h = \frac{4 h^{2}}{4} + \frac{8}{4} + \frac{4 h}{4}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Cancel the common factor of $4$ .

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Step 4.3.1.1.1

Cancel the common factor.

$h = \frac{4 h^{2}}{4} + \frac{8}{4} + \frac{4 h}{4}$

Step 4.3.1.1.2

Divide $h^{2}$ by $1$ .

$h = h^{2} + \frac{8}{4} + \frac{4 h}{4}$

Step 4.3.1.2

Divide $8$ by $4$ .

$h = h^{2} + 2 + \frac{4 h}{4}$

Step 4.3.1.3

Cancel the common factor of $4$ .

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Step 4.3.1.3.1

Cancel the common factor.

$h = h^{2} + 2 + \frac{4 h}{4}$

Step 4.3.1.3.2

Divide $h$ by $1$ .

$h = h^{2} + 2 + h$

Step 5

The variable $x$ got canceled.

All real numbers

Step 6

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$